Whether the statement “If r ( t ) = 〈 cos t , e t , t 〉 + C and r ( 0 ) = 〈 0 , 0 , 0 〉 , then C = 〈 0 , 0 , 0 〉 ” is true or not.

Question

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Answer 1

Question

Chapter 14, Problem 1RE

a.

To determine

Whether the statement “If $r(t)=〈cost,et,t〉+C$ and $r(0)=〈0,0,0〉$ , then $C=〈0,0,0〉$ ” is true or not.

a.

Expert Solution

Answer to Problem 1RE

The statement is false.

Explanation of Solution

The given vector valued function is $r(t)=〈cost,et,t〉+C$ .

Substitute $t=0$ in the vector as follows.

$r(0)=〈cos(0),e(0),(0)〉+C〈0,0,0〉=〈1,1,0〉+CC=〈0,0,0〉−〈1,1,0〉 =〈−1,−1,0〉$

Thus, the constant vector is $C=〈−1,−1,0〉$ .

Hence the statement if $r(t)=〈cost,et,t〉+C$ and $r(0)=〈0,0,0〉$ , then $C=〈0,0,0〉$ is false.

b.

To determine

Whether the statement “The curvature of a circle of radius 5 is $κ=15$ ” is true or not.

b.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Definition used:

“The curvature of a curve at a point can be visualized in terms of a circle of curvature, which is a circle of radius R that is tangent to the curve at that point.

The curvature at the point is $κ=1R$ ”

Description:

The given circle has the radius of 5.

By the above definition used, the curvature of a circle is reciprocal of the radius of the circle.

Thus, the curvature of the circle is $κ=15$ .

Hence, the statement “the curvature of a circle of radius 5 is $κ=15$ ” is true.

c.

To determine

Whether the statement “The graph of the curve $r(t)=〈3cost,0,6sint〉$ is an ellipse in the xz-plane” is or not.

c.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Definition used:

One point to a curve corresponds to a single vector $r=〈x,y,z〉$ where $x=x0+at, y=y0+bt and z=z0+ct$ .

The entire curve can be represented by a vector-valued function $r(t)=〈x(t),y(t),z(t)〉$ .

Calculation:

The given graph is $r(t)=〈3cost,0,6sint〉$ .

The values are $〈x(t),y(t),z(t)〉=〈3cost,0,6sint〉$ .

Here, $〈3cost,6sint〉$ can represent an ellipse in x and z axis.

That is $x232+z262=1$ .

Therefore, the statement “the graph of the curve $r(t)=〈3cost,0,6sint〉$ is an ellipse in the xz-plane” is true.

d.

To determine

Whether the given statement “If $r′(t)=0, then r(t)=〈a,b,c〉$ , where a, b, and c are real numbers.” is true or not.

d.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Consider $r′(t)=0$ and integrate with respect to t.

$r(t)=∫r′(t)dt=∫〈0,0,0〉dt=〈0,0,0〉+C=〈0,0,0〉+〈a,b,c〉 (∵C=〈a,b,c〉)=〈a,b,c〉$

Thus, the vector $r(t)$ is $〈a,b,c〉$ where a, b, and c are real numbers.

Therefore, the given statement is true.

e.

To determine

Whether the statement “The parameterized curve $r(t)=〈5cost,12cost,13sint〉$ has arc length as a parameter.” is true or not.

e.

Expert Solution

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Suppose $r(t)$ is a smooth curve. If $|v(t)|=1$ , then the curve uses arc length as a parameter.

Differentiate $r(t)$ to compute $v(t)$ .

$v(t)=〈(5cost)′,(12cost)′,(13sint)′〉=〈−5sint,−12sint,−13cost〉$

Compute $|v(t)|$ .

$|v(t)|=(−5sint)2+(−12sint)2+(−13cost)2=25sin2t+144sin2t+169cos2t=169(sin2t+cos2t)=169=13$

Since $|v(t)|≠1$ , the curve does not use arc length as a parameter.

Therefore, the given statement is false.

f.

To determine

Whether the statement “The position vector and the principal unit normal are always parallel on a smooth curve.” is true or not.

f.

Expert Solution

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Formula used:

Suppose r is a smooth parameterized curve and s is the arc length.

The unit tangent vector T is $r′|r′|$ .

The principal unit normal vector is $N(t)=T′(t)|T′(t)|$ .

Counter example:

Consider $r(t)=〈4sint,4cost,10t〉$ .

Differentiate $r(t)$ to compute $r′(t)$ .

$r′(t)=〈(4sint)′,(4cost)′,(10t)′〉=〈4cost,−4sint,10〉$

Use magnitude formula to obtain the value of $|r′(t)|$ .

$|r′(t)|=|〈4cost,−4sint,10〉|=(4cost)2+(−4sint)2+102=16cos2t+16sin2t+100=16(cos2t+sin2t)+100$

On further simplification,

$|r′(t)|=16(1)+100=116=229$

Use unit tangent formula to compute $T(t)$ .

$T(t)=〈4cost,−4sint,10〉229=129〈4cost2,−4sint2,102〉=129〈2cost,−2sint,5〉$

Thus, the unit tangent vector $T(t)$ is $129〈2cost,−2sint,5〉$ .

Differentiate $T(t)$ to compute $T′(t)$ .

$T′(t)=129〈(2cost)′,(−2sint)′,5′〉=129〈−2sint,−2cost,0〉$

Use magnitude formula to obtain the value of $|T′(t)|$ .

$|T′(t)|=|129〈−2sint,−2cost,0〉|=129(−2sint)2+(−2cost)2+02=1294sin2t+4cos2t=1294(sin2t+cos2t)$

On further simplification,

$|T′(t)|=1294(1)=229$

Use principal unit normal formula to compute the value of $N(t)$ .

$N(t)=129〈−2sint,−2cost,0〉229=129(292)〈−2sint,−2cost,0〉=12〈−2sint,−2cost,0〉=〈−sint2,−cost2,02〉$

$=〈−sint,−cost,0〉$

Thus, the principal unit normal vector $N(t)$ for the curve $r(t)$ is $〈−sint,−cost,0〉$ .

It is observed that the position vector and the principal unit normal vector are not equal.

Therefore, the given statement is false.

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Answer 2

Question

Chapter 14, Problem 1RE

a.

To determine

Whether the statement “If $r(t)=〈cost,et,t〉+C$ and $r(0)=〈0,0,0〉$ , then $C=〈0,0,0〉$ ” is true or not.

a.

Expert Solution

Answer to Problem 1RE

The statement is false.

Explanation of Solution

The given vector valued function is $r(t)=〈cost,et,t〉+C$ .

Substitute $t=0$ in the vector as follows.

$r(0)=〈cos(0),e(0),(0)〉+C〈0,0,0〉=〈1,1,0〉+CC=〈0,0,0〉−〈1,1,0〉 =〈−1,−1,0〉$

Thus, the constant vector is $C=〈−1,−1,0〉$ .

Hence the statement if $r(t)=〈cost,et,t〉+C$ and $r(0)=〈0,0,0〉$ , then $C=〈0,0,0〉$ is false.

b.

To determine

Whether the statement “The curvature of a circle of radius 5 is $κ=15$ ” is true or not.

b.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Definition used:

“The curvature of a curve at a point can be visualized in terms of a circle of curvature, which is a circle of radius R that is tangent to the curve at that point.

The curvature at the point is $κ=1R$ ”

Description:

The given circle has the radius of 5.

By the above definition used, the curvature of a circle is reciprocal of the radius of the circle.

Thus, the curvature of the circle is $κ=15$ .

Hence, the statement “the curvature of a circle of radius 5 is $κ=15$ ” is true.

c.

To determine

Whether the statement “The graph of the curve $r(t)=〈3cost,0,6sint〉$ is an ellipse in the xz-plane” is or not.

c.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Definition used:

One point to a curve corresponds to a single vector $r=〈x,y,z〉$ where $x=x0+at, y=y0+bt and z=z0+ct$ .

The entire curve can be represented by a vector-valued function $r(t)=〈x(t),y(t),z(t)〉$ .

Calculation:

The given graph is $r(t)=〈3cost,0,6sint〉$ .

The values are $〈x(t),y(t),z(t)〉=〈3cost,0,6sint〉$ .

Here, $〈3cost,6sint〉$ can represent an ellipse in x and z axis.

That is $x232+z262=1$ .

Therefore, the statement “the graph of the curve $r(t)=〈3cost,0,6sint〉$ is an ellipse in the xz-plane” is true.

d.

To determine

Whether the given statement “If $r′(t)=0, then r(t)=〈a,b,c〉$ , where a, b, and c are real numbers.” is true or not.

d.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Consider $r′(t)=0$ and integrate with respect to t.

$r(t)=∫r′(t)dt=∫〈0,0,0〉dt=〈0,0,0〉+C=〈0,0,0〉+〈a,b,c〉 (∵C=〈a,b,c〉)=〈a,b,c〉$

Thus, the vector $r(t)$ is $〈a,b,c〉$ where a, b, and c are real numbers.

Therefore, the given statement is true.

e.

To determine

Whether the statement “The parameterized curve $r(t)=〈5cost,12cost,13sint〉$ has arc length as a parameter.” is true or not.

e.

Expert Solution

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Suppose $r(t)$ is a smooth curve. If $|v(t)|=1$ , then the curve uses arc length as a parameter.

Differentiate $r(t)$ to compute $v(t)$ .

$v(t)=〈(5cost)′,(12cost)′,(13sint)′〉=〈−5sint,−12sint,−13cost〉$

Compute $|v(t)|$ .

$|v(t)|=(−5sint)2+(−12sint)2+(−13cost)2=25sin2t+144sin2t+169cos2t=169(sin2t+cos2t)=169=13$

Since $|v(t)|≠1$ , the curve does not use arc length as a parameter.

Therefore, the given statement is false.

f.

To determine

Whether the statement “The position vector and the principal unit normal are always parallel on a smooth curve.” is true or not.

f.

Expert Solution

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Formula used:

Suppose r is a smooth parameterized curve and s is the arc length.

The unit tangent vector T is $r′|r′|$ .

The principal unit normal vector is $N(t)=T′(t)|T′(t)|$ .

Counter example:

Consider $r(t)=〈4sint,4cost,10t〉$ .

Differentiate $r(t)$ to compute $r′(t)$ .

$r′(t)=〈(4sint)′,(4cost)′,(10t)′〉=〈4cost,−4sint,10〉$

Use magnitude formula to obtain the value of $|r′(t)|$ .

$|r′(t)|=|〈4cost,−4sint,10〉|=(4cost)2+(−4sint)2+102=16cos2t+16sin2t+100=16(cos2t+sin2t)+100$

On further simplification,

$|r′(t)|=16(1)+100=116=229$

Use unit tangent formula to compute $T(t)$ .

$T(t)=〈4cost,−4sint,10〉229=129〈4cost2,−4sint2,102〉=129〈2cost,−2sint,5〉$

Thus, the unit tangent vector $T(t)$ is $129〈2cost,−2sint,5〉$ .

Differentiate $T(t)$ to compute $T′(t)$ .

$T′(t)=129〈(2cost)′,(−2sint)′,5′〉=129〈−2sint,−2cost,0〉$

Use magnitude formula to obtain the value of $|T′(t)|$ .

$|T′(t)|=|129〈−2sint,−2cost,0〉|=129(−2sint)2+(−2cost)2+02=1294sin2t+4cos2t=1294(sin2t+cos2t)$

On further simplification,

$|T′(t)|=1294(1)=229$

Use principal unit normal formula to compute the value of $N(t)$ .

$N(t)=129〈−2sint,−2cost,0〉229=129(292)〈−2sint,−2cost,0〉=12〈−2sint,−2cost,0〉=〈−sint2,−cost2,02〉$

$=〈−sint,−cost,0〉$

Thus, the principal unit normal vector $N(t)$ for the curve $r(t)$ is $〈−sint,−cost,0〉$ .

It is observed that the position vector and the principal unit normal vector are not equal.

Therefore, the given statement is false.

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Answer 3

Question

Chapter 14, Problem 1RE

a.

To determine

Whether the statement “If $r(t)=〈cost,et,t〉+C$ and $r(0)=〈0,0,0〉$ , then $C=〈0,0,0〉$ ” is true or not.

a.

Expert Solution

Answer to Problem 1RE

The statement is false.

Explanation of Solution

The given vector valued function is $r(t)=〈cost,et,t〉+C$ .

Substitute $t=0$ in the vector as follows.

$r(0)=〈cos(0),e(0),(0)〉+C〈0,0,0〉=〈1,1,0〉+CC=〈0,0,0〉−〈1,1,0〉 =〈−1,−1,0〉$

Thus, the constant vector is $C=〈−1,−1,0〉$ .

Hence the statement if $r(t)=〈cost,et,t〉+C$ and $r(0)=〈0,0,0〉$ , then $C=〈0,0,0〉$ is false.

b.

To determine

Whether the statement “The curvature of a circle of radius 5 is $κ=15$ ” is true or not.

b.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Definition used:

“The curvature of a curve at a point can be visualized in terms of a circle of curvature, which is a circle of radius R that is tangent to the curve at that point.

The curvature at the point is $κ=1R$ ”

Description:

The given circle has the radius of 5.

By the above definition used, the curvature of a circle is reciprocal of the radius of the circle.

Thus, the curvature of the circle is $κ=15$ .

Hence, the statement “the curvature of a circle of radius 5 is $κ=15$ ” is true.

c.

To determine

Whether the statement “The graph of the curve $r(t)=〈3cost,0,6sint〉$ is an ellipse in the xz-plane” is or not.

c.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Definition used:

One point to a curve corresponds to a single vector $r=〈x,y,z〉$ where $x=x0+at, y=y0+bt and z=z0+ct$ .

The entire curve can be represented by a vector-valued function $r(t)=〈x(t),y(t),z(t)〉$ .

Calculation:

The given graph is $r(t)=〈3cost,0,6sint〉$ .

The values are $〈x(t),y(t),z(t)〉=〈3cost,0,6sint〉$ .

Here, $〈3cost,6sint〉$ can represent an ellipse in x and z axis.

That is $x232+z262=1$ .

Therefore, the statement “the graph of the curve $r(t)=〈3cost,0,6sint〉$ is an ellipse in the xz-plane” is true.

d.

To determine

Whether the given statement “If $r′(t)=0, then r(t)=〈a,b,c〉$ , where a, b, and c are real numbers.” is true or not.

d.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Consider $r′(t)=0$ and integrate with respect to t.

$r(t)=∫r′(t)dt=∫〈0,0,0〉dt=〈0,0,0〉+C=〈0,0,0〉+〈a,b,c〉 (∵C=〈a,b,c〉)=〈a,b,c〉$

Thus, the vector $r(t)$ is $〈a,b,c〉$ where a, b, and c are real numbers.

Therefore, the given statement is true.

e.

To determine

Whether the statement “The parameterized curve $r(t)=〈5cost,12cost,13sint〉$ has arc length as a parameter.” is true or not.

e.

Expert Solution

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Suppose $r(t)$ is a smooth curve. If $|v(t)|=1$ , then the curve uses arc length as a parameter.

Differentiate $r(t)$ to compute $v(t)$ .

$v(t)=〈(5cost)′,(12cost)′,(13sint)′〉=〈−5sint,−12sint,−13cost〉$

Compute $|v(t)|$ .

$|v(t)|=(−5sint)2+(−12sint)2+(−13cost)2=25sin2t+144sin2t+169cos2t=169(sin2t+cos2t)=169=13$

Since $|v(t)|≠1$ , the curve does not use arc length as a parameter.

Therefore, the given statement is false.

f.

To determine

Whether the statement “The position vector and the principal unit normal are always parallel on a smooth curve.” is true or not.

f.

Expert Solution

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Formula used:

Suppose r is a smooth parameterized curve and s is the arc length.

The unit tangent vector T is $r′|r′|$ .

The principal unit normal vector is $N(t)=T′(t)|T′(t)|$ .

Counter example:

Consider $r(t)=〈4sint,4cost,10t〉$ .

Differentiate $r(t)$ to compute $r′(t)$ .

$r′(t)=〈(4sint)′,(4cost)′,(10t)′〉=〈4cost,−4sint,10〉$

Use magnitude formula to obtain the value of $|r′(t)|$ .

$|r′(t)|=|〈4cost,−4sint,10〉|=(4cost)2+(−4sint)2+102=16cos2t+16sin2t+100=16(cos2t+sin2t)+100$

On further simplification,

$|r′(t)|=16(1)+100=116=229$

Use unit tangent formula to compute $T(t)$ .

$T(t)=〈4cost,−4sint,10〉229=129〈4cost2,−4sint2,102〉=129〈2cost,−2sint,5〉$

Thus, the unit tangent vector $T(t)$ is $129〈2cost,−2sint,5〉$ .

Differentiate $T(t)$ to compute $T′(t)$ .

$T′(t)=129〈(2cost)′,(−2sint)′,5′〉=129〈−2sint,−2cost,0〉$

Use magnitude formula to obtain the value of $|T′(t)|$ .

$|T′(t)|=|129〈−2sint,−2cost,0〉|=129(−2sint)2+(−2cost)2+02=1294sin2t+4cos2t=1294(sin2t+cos2t)$

On further simplification,

$|T′(t)|=1294(1)=229$

Use principal unit normal formula to compute the value of $N(t)$ .

$N(t)=129〈−2sint,−2cost,0〉229=129(292)〈−2sint,−2cost,0〉=12〈−2sint,−2cost,0〉=〈−sint2,−cost2,02〉$

$=〈−sint,−cost,0〉$

Thus, the principal unit normal vector $N(t)$ for the curve $r(t)$ is $〈−sint,−cost,0〉$ .

It is observed that the position vector and the principal unit normal vector are not equal.

Therefore, the given statement is false.

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Answer 4

Question

Chapter 14, Problem 1RE

a.

To determine

Whether the statement “If $r(t)=〈cost,et,t〉+C$ and $r(0)=〈0,0,0〉$ , then $C=〈0,0,0〉$ ” is true or not.

a.

Expert Solution

Answer to Problem 1RE

The statement is false.

Explanation of Solution

The given vector valued function is $r(t)=〈cost,et,t〉+C$ .

Substitute $t=0$ in the vector as follows.

$r(0)=〈cos(0),e(0),(0)〉+C〈0,0,0〉=〈1,1,0〉+CC=〈0,0,0〉−〈1,1,0〉 =〈−1,−1,0〉$

Thus, the constant vector is $C=〈−1,−1,0〉$ .

Hence the statement if $r(t)=〈cost,et,t〉+C$ and $r(0)=〈0,0,0〉$ , then $C=〈0,0,0〉$ is false.

b.

To determine

Whether the statement “The curvature of a circle of radius 5 is $κ=15$ ” is true or not.

b.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Definition used:

“The curvature of a curve at a point can be visualized in terms of a circle of curvature, which is a circle of radius R that is tangent to the curve at that point.

The curvature at the point is $κ=1R$ ”

Description:

The given circle has the radius of 5.

By the above definition used, the curvature of a circle is reciprocal of the radius of the circle.

Thus, the curvature of the circle is $κ=15$ .

Hence, the statement “the curvature of a circle of radius 5 is $κ=15$ ” is true.

c.

To determine

Whether the statement “The graph of the curve $r(t)=〈3cost,0,6sint〉$ is an ellipse in the xz-plane” is or not.

c.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Definition used:

One point to a curve corresponds to a single vector $r=〈x,y,z〉$ where $x=x0+at, y=y0+bt and z=z0+ct$ .

The entire curve can be represented by a vector-valued function $r(t)=〈x(t),y(t),z(t)〉$ .

Calculation:

The given graph is $r(t)=〈3cost,0,6sint〉$ .

The values are $〈x(t),y(t),z(t)〉=〈3cost,0,6sint〉$ .

Here, $〈3cost,6sint〉$ can represent an ellipse in x and z axis.

That is $x232+z262=1$ .

Therefore, the statement “the graph of the curve $r(t)=〈3cost,0,6sint〉$ is an ellipse in the xz-plane” is true.

d.

To determine

Whether the given statement “If $r′(t)=0, then r(t)=〈a,b,c〉$ , where a, b, and c are real numbers.” is true or not.

d.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Consider $r′(t)=0$ and integrate with respect to t.

$r(t)=∫r′(t)dt=∫〈0,0,0〉dt=〈0,0,0〉+C=〈0,0,0〉+〈a,b,c〉 (∵C=〈a,b,c〉)=〈a,b,c〉$

Thus, the vector $r(t)$ is $〈a,b,c〉$ where a, b, and c are real numbers.

Therefore, the given statement is true.

e.

To determine

Whether the statement “The parameterized curve $r(t)=〈5cost,12cost,13sint〉$ has arc length as a parameter.” is true or not.

e.

Expert Solution

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Suppose $r(t)$ is a smooth curve. If $|v(t)|=1$ , then the curve uses arc length as a parameter.

Differentiate $r(t)$ to compute $v(t)$ .

$v(t)=〈(5cost)′,(12cost)′,(13sint)′〉=〈−5sint,−12sint,−13cost〉$

Compute $|v(t)|$ .

$|v(t)|=(−5sint)2+(−12sint)2+(−13cost)2=25sin2t+144sin2t+169cos2t=169(sin2t+cos2t)=169=13$

Since $|v(t)|≠1$ , the curve does not use arc length as a parameter.

Therefore, the given statement is false.

f.

To determine

Whether the statement “The position vector and the principal unit normal are always parallel on a smooth curve.” is true or not.

f.

Expert Solution

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Formula used:

Suppose r is a smooth parameterized curve and s is the arc length.

The unit tangent vector T is $r′|r′|$ .

The principal unit normal vector is $N(t)=T′(t)|T′(t)|$ .

Counter example:

Consider $r(t)=〈4sint,4cost,10t〉$ .

Differentiate $r(t)$ to compute $r′(t)$ .

$r′(t)=〈(4sint)′,(4cost)′,(10t)′〉=〈4cost,−4sint,10〉$

Use magnitude formula to obtain the value of $|r′(t)|$ .

$|r′(t)|=|〈4cost,−4sint,10〉|=(4cost)2+(−4sint)2+102=16cos2t+16sin2t+100=16(cos2t+sin2t)+100$

On further simplification,

$|r′(t)|=16(1)+100=116=229$

Use unit tangent formula to compute $T(t)$ .

$T(t)=〈4cost,−4sint,10〉229=129〈4cost2,−4sint2,102〉=129〈2cost,−2sint,5〉$

Thus, the unit tangent vector $T(t)$ is $129〈2cost,−2sint,5〉$ .

Differentiate $T(t)$ to compute $T′(t)$ .

$T′(t)=129〈(2cost)′,(−2sint)′,5′〉=129〈−2sint,−2cost,0〉$

Use magnitude formula to obtain the value of $|T′(t)|$ .

$|T′(t)|=|129〈−2sint,−2cost,0〉|=129(−2sint)2+(−2cost)2+02=1294sin2t+4cos2t=1294(sin2t+cos2t)$

On further simplification,

$|T′(t)|=1294(1)=229$

Use principal unit normal formula to compute the value of $N(t)$ .

$N(t)=129〈−2sint,−2cost,0〉229=129(292)〈−2sint,−2cost,0〉=12〈−2sint,−2cost,0〉=〈−sint2,−cost2,02〉$

$=〈−sint,−cost,0〉$

Thus, the principal unit normal vector $N(t)$ for the curve $r(t)$ is $〈−sint,−cost,0〉$ .

It is observed that the position vector and the principal unit normal vector are not equal.

Therefore, the given statement is false.

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Answer 5

Chapter 14, Problem 1RE

a.

To determine

Whether the statement “If $r(t)=〈cost,et,t〉+C$ and $r(0)=〈0,0,0〉$ , then $C=〈0,0,0〉$ ” is true or not.

a.

Expert Solution

Answer to Problem 1RE

The statement is false.

Explanation of Solution

The given vector valued function is $r(t)=〈cost,et,t〉+C$ .

Substitute $t=0$ in the vector as follows.

$r(0)=〈cos(0),e(0),(0)〉+C〈0,0,0〉=〈1,1,0〉+CC=〈0,0,0〉−〈1,1,0〉 =〈−1,−1,0〉$

Thus, the constant vector is $C=〈−1,−1,0〉$ .

Hence the statement if $r(t)=〈cost,et,t〉+C$ and $r(0)=〈0,0,0〉$ , then $C=〈0,0,0〉$ is false.

b.

To determine

Whether the statement “The curvature of a circle of radius 5 is $κ=15$ ” is true or not.

b.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Definition used:

“The curvature of a curve at a point can be visualized in terms of a circle of curvature, which is a circle of radius R that is tangent to the curve at that point.

The curvature at the point is $κ=1R$ ”

Description:

The given circle has the radius of 5.

By the above definition used, the curvature of a circle is reciprocal of the radius of the circle.

Thus, the curvature of the circle is $κ=15$ .

Hence, the statement “the curvature of a circle of radius 5 is $κ=15$ ” is true.

c.

To determine

Whether the statement “The graph of the curve $r(t)=〈3cost,0,6sint〉$ is an ellipse in the xz-plane” is or not.

c.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Definition used:

One point to a curve corresponds to a single vector $r=〈x,y,z〉$ where $x=x0+at, y=y0+bt and z=z0+ct$ .

The entire curve can be represented by a vector-valued function $r(t)=〈x(t),y(t),z(t)〉$ .

Calculation:

The given graph is $r(t)=〈3cost,0,6sint〉$ .

The values are $〈x(t),y(t),z(t)〉=〈3cost,0,6sint〉$ .

Here, $〈3cost,6sint〉$ can represent an ellipse in x and z axis.

That is $x232+z262=1$ .

Therefore, the statement “the graph of the curve $r(t)=〈3cost,0,6sint〉$ is an ellipse in the xz-plane” is true.

d.

To determine

Whether the given statement “If $r′(t)=0, then r(t)=〈a,b,c〉$ , where a, b, and c are real numbers.” is true or not.

d.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Consider $r′(t)=0$ and integrate with respect to t.

$r(t)=∫r′(t)dt=∫〈0,0,0〉dt=〈0,0,0〉+C=〈0,0,0〉+〈a,b,c〉 (∵C=〈a,b,c〉)=〈a,b,c〉$

Thus, the vector $r(t)$ is $〈a,b,c〉$ where a, b, and c are real numbers.

Therefore, the given statement is true.

e.

To determine

Whether the statement “The parameterized curve $r(t)=〈5cost,12cost,13sint〉$ has arc length as a parameter.” is true or not.

e.

Expert Solution

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Suppose $r(t)$ is a smooth curve. If $|v(t)|=1$ , then the curve uses arc length as a parameter.

Differentiate $r(t)$ to compute $v(t)$ .

$v(t)=〈(5cost)′,(12cost)′,(13sint)′〉=〈−5sint,−12sint,−13cost〉$

Compute $|v(t)|$ .

$|v(t)|=(−5sint)2+(−12sint)2+(−13cost)2=25sin2t+144sin2t+169cos2t=169(sin2t+cos2t)=169=13$

Since $|v(t)|≠1$ , the curve does not use arc length as a parameter.

Therefore, the given statement is false.

f.

To determine

Whether the statement “The position vector and the principal unit normal are always parallel on a smooth curve.” is true or not.

f.

Expert Solution

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Formula used:

Suppose r is a smooth parameterized curve and s is the arc length.

The unit tangent vector T is $r′|r′|$ .

The principal unit normal vector is $N(t)=T′(t)|T′(t)|$ .

Counter example:

Consider $r(t)=〈4sint,4cost,10t〉$ .

Differentiate $r(t)$ to compute $r′(t)$ .

$r′(t)=〈(4sint)′,(4cost)′,(10t)′〉=〈4cost,−4sint,10〉$

Use magnitude formula to obtain the value of $|r′(t)|$ .

$|r′(t)|=|〈4cost,−4sint,10〉|=(4cost)2+(−4sint)2+102=16cos2t+16sin2t+100=16(cos2t+sin2t)+100$

On further simplification,

$|r′(t)|=16(1)+100=116=229$

Use unit tangent formula to compute $T(t)$ .

$T(t)=〈4cost,−4sint,10〉229=129〈4cost2,−4sint2,102〉=129〈2cost,−2sint,5〉$

Thus, the unit tangent vector $T(t)$ is $129〈2cost,−2sint,5〉$ .

Differentiate $T(t)$ to compute $T′(t)$ .

$T′(t)=129〈(2cost)′,(−2sint)′,5′〉=129〈−2sint,−2cost,0〉$

Use magnitude formula to obtain the value of $|T′(t)|$ .

$|T′(t)|=|129〈−2sint,−2cost,0〉|=129(−2sint)2+(−2cost)2+02=1294sin2t+4cos2t=1294(sin2t+cos2t)$

On further simplification,

$|T′(t)|=1294(1)=229$

Use principal unit normal formula to compute the value of $N(t)$ .

$N(t)=129〈−2sint,−2cost,0〉229=129(292)〈−2sint,−2cost,0〉=12〈−2sint,−2cost,0〉=〈−sint2,−cost2,02〉$

$=〈−sint,−cost,0〉$

Thus, the principal unit normal vector $N(t)$ for the curve $r(t)$ is $〈−sint,−cost,0〉$ .

It is observed that the position vector and the principal unit normal vector are not equal.

Therefore, the given statement is false.

Answer 6

Chapter 14, Problem 1RE

a.

To determine

Whether the statement “If $r(t)=〈cost,et,t〉+C$ and $r(0)=〈0,0,0〉$ , then $C=〈0,0,0〉$ ” is true or not.

a.

Expert Solution

Answer to Problem 1RE

The statement is false.

Explanation of Solution

The given vector valued function is $r(t)=〈cost,et,t〉+C$ .

Substitute $t=0$ in the vector as follows.

$r(0)=〈cos(0),e(0),(0)〉+C〈0,0,0〉=〈1,1,0〉+CC=〈0,0,0〉−〈1,1,0〉 =〈−1,−1,0〉$

Thus, the constant vector is $C=〈−1,−1,0〉$ .

Hence the statement if $r(t)=〈cost,et,t〉+C$ and $r(0)=〈0,0,0〉$ , then $C=〈0,0,0〉$ is false.

Answer 7

Chapter 14, Problem 1RE

a.

To determine

Whether the statement “If $r(t)=〈cost,et,t〉+C$ and $r(0)=〈0,0,0〉$ , then $C=〈0,0,0〉$ ” is true or not.

Answer 8

a.

Expert Solution

Answer to Problem 1RE

The statement is false.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The given vector valued function is $r(t)=〈cost,et,t〉+C$ .

Substitute $t=0$ in the vector as follows.

$r(0)=〈cos(0),e(0),(0)〉+C〈0,0,0〉=〈1,1,0〉+CC=〈0,0,0〉−〈1,1,0〉 =〈−1,−1,0〉$

Thus, the constant vector is $C=〈−1,−1,0〉$ .

Hence the statement if $r(t)=〈cost,et,t〉+C$ and $r(0)=〈0,0,0〉$ , then $C=〈0,0,0〉$ is false.

Answer 13

b.

To determine

Whether the statement “The curvature of a circle of radius 5 is $κ=15$ ” is true or not.

b.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Definition used:

“The curvature of a curve at a point can be visualized in terms of a circle of curvature, which is a circle of radius R that is tangent to the curve at that point.

The curvature at the point is $κ=1R$ ”

Description:

The given circle has the radius of 5.

By the above definition used, the curvature of a circle is reciprocal of the radius of the circle.

Thus, the curvature of the circle is $κ=15$ .

Hence, the statement “the curvature of a circle of radius 5 is $κ=15$ ” is true.

Answer 14

b.

To determine

Whether the statement “The curvature of a circle of radius 5 is $κ=15$ ” is true or not.

Answer 15

b.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Definition used:

“The curvature of a curve at a point can be visualized in terms of a circle of curvature, which is a circle of radius R that is tangent to the curve at that point.

The curvature at the point is $κ=1R$ ”

Description:

The given circle has the radius of 5.

By the above definition used, the curvature of a circle is reciprocal of the radius of the circle.

Thus, the curvature of the circle is $κ=15$ .

Hence, the statement “the curvature of a circle of radius 5 is $κ=15$ ” is true.

Answer 20

c.

To determine

Whether the statement “The graph of the curve $r(t)=〈3cost,0,6sint〉$ is an ellipse in the xz-plane” is or not.

c.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Definition used:

One point to a curve corresponds to a single vector $r=〈x,y,z〉$ where $x=x0+at, y=y0+bt and z=z0+ct$ .

The entire curve can be represented by a vector-valued function $r(t)=〈x(t),y(t),z(t)〉$ .

Calculation:

The given graph is $r(t)=〈3cost,0,6sint〉$ .

The values are $〈x(t),y(t),z(t)〉=〈3cost,0,6sint〉$ .

Here, $〈3cost,6sint〉$ can represent an ellipse in x and z axis.

That is $x232+z262=1$ .

Therefore, the statement “the graph of the curve $r(t)=〈3cost,0,6sint〉$ is an ellipse in the xz-plane” is true.

Answer 21

c.

To determine

Whether the statement “The graph of the curve $r(t)=〈3cost,0,6sint〉$ is an ellipse in the xz-plane” is or not.

Answer 22

c.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Definition used:

One point to a curve corresponds to a single vector $r=〈x,y,z〉$ where $x=x0+at, y=y0+bt and z=z0+ct$ .

The entire curve can be represented by a vector-valued function $r(t)=〈x(t),y(t),z(t)〉$ .

Calculation:

The given graph is $r(t)=〈3cost,0,6sint〉$ .

The values are $〈x(t),y(t),z(t)〉=〈3cost,0,6sint〉$ .

Here, $〈3cost,6sint〉$ can represent an ellipse in x and z axis.

That is $x232+z262=1$ .

Therefore, the statement “the graph of the curve $r(t)=〈3cost,0,6sint〉$ is an ellipse in the xz-plane” is true.

Answer 27

d.

To determine

Whether the given statement “If $r′(t)=0, then r(t)=〈a,b,c〉$ , where a, b, and c are real numbers.” is true or not.

d.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Consider $r′(t)=0$ and integrate with respect to t.

$r(t)=∫r′(t)dt=∫〈0,0,0〉dt=〈0,0,0〉+C=〈0,0,0〉+〈a,b,c〉 (∵C=〈a,b,c〉)=〈a,b,c〉$

Thus, the vector $r(t)$ is $〈a,b,c〉$ where a, b, and c are real numbers.

Therefore, the given statement is true.

Answer 28

d.

To determine

Whether the given statement “If $r′(t)=0, then r(t)=〈a,b,c〉$ , where a, b, and c are real numbers.” is true or not.

Answer 29

d.

Expert Solution

Answer to Problem 1RE

The given statement is true.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Consider $r′(t)=0$ and integrate with respect to t.

$r(t)=∫r′(t)dt=∫〈0,0,0〉dt=〈0,0,0〉+C=〈0,0,0〉+〈a,b,c〉 (∵C=〈a,b,c〉)=〈a,b,c〉$

Thus, the vector $r(t)$ is $〈a,b,c〉$ where a, b, and c are real numbers.

Therefore, the given statement is true.

Answer 34

e.

To determine

Whether the statement “The parameterized curve $r(t)=〈5cost,12cost,13sint〉$ has arc length as a parameter.” is true or not.

e.

Expert Solution

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Suppose $r(t)$ is a smooth curve. If $|v(t)|=1$ , then the curve uses arc length as a parameter.

Differentiate $r(t)$ to compute $v(t)$ .

$v(t)=〈(5cost)′,(12cost)′,(13sint)′〉=〈−5sint,−12sint,−13cost〉$

Compute $|v(t)|$ .

$|v(t)|=(−5sint)2+(−12sint)2+(−13cost)2=25sin2t+144sin2t+169cos2t=169(sin2t+cos2t)=169=13$

Since $|v(t)|≠1$ , the curve does not use arc length as a parameter.

Therefore, the given statement is false.

Answer 35

e.

To determine

Whether the statement “The parameterized curve $r(t)=〈5cost,12cost,13sint〉$ has arc length as a parameter.” is true or not.

Answer 36

e.

Expert Solution

Answer to Problem 1RE

The given statement is false.

Answer 37

e.

Expert Solution

Answer 38

e.

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Suppose $r(t)$ is a smooth curve. If $|v(t)|=1$ , then the curve uses arc length as a parameter.

Differentiate $r(t)$ to compute $v(t)$ .

$v(t)=〈(5cost)′,(12cost)′,(13sint)′〉=〈−5sint,−12sint,−13cost〉$

Compute $|v(t)|$ .

$|v(t)|=(−5sint)2+(−12sint)2+(−13cost)2=25sin2t+144sin2t+169cos2t=169(sin2t+cos2t)=169=13$

Since $|v(t)|≠1$ , the curve does not use arc length as a parameter.

Therefore, the given statement is false.

Answer 41

f.

To determine

Whether the statement “The position vector and the principal unit normal are always parallel on a smooth curve.” is true or not.

f.

Expert Solution

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Formula used:

Suppose r is a smooth parameterized curve and s is the arc length.

The unit tangent vector T is $r′|r′|$ .

The principal unit normal vector is $N(t)=T′(t)|T′(t)|$ .

Counter example:

Consider $r(t)=〈4sint,4cost,10t〉$ .

Differentiate $r(t)$ to compute $r′(t)$ .

$r′(t)=〈(4sint)′,(4cost)′,(10t)′〉=〈4cost,−4sint,10〉$

Use magnitude formula to obtain the value of $|r′(t)|$ .

$|r′(t)|=|〈4cost,−4sint,10〉|=(4cost)2+(−4sint)2+102=16cos2t+16sin2t+100=16(cos2t+sin2t)+100$

On further simplification,

$|r′(t)|=16(1)+100=116=229$

Use unit tangent formula to compute $T(t)$ .

$T(t)=〈4cost,−4sint,10〉229=129〈4cost2,−4sint2,102〉=129〈2cost,−2sint,5〉$

Thus, the unit tangent vector $T(t)$ is $129〈2cost,−2sint,5〉$ .

Differentiate $T(t)$ to compute $T′(t)$ .

$T′(t)=129〈(2cost)′,(−2sint)′,5′〉=129〈−2sint,−2cost,0〉$

Use magnitude formula to obtain the value of $|T′(t)|$ .

$|T′(t)|=|129〈−2sint,−2cost,0〉|=129(−2sint)2+(−2cost)2+02=1294sin2t+4cos2t=1294(sin2t+cos2t)$

On further simplification,

$|T′(t)|=1294(1)=229$

Use principal unit normal formula to compute the value of $N(t)$ .

$N(t)=129〈−2sint,−2cost,0〉229=129(292)〈−2sint,−2cost,0〉=12〈−2sint,−2cost,0〉=〈−sint2,−cost2,02〉$

$=〈−sint,−cost,0〉$

Thus, the principal unit normal vector $N(t)$ for the curve $r(t)$ is $〈−sint,−cost,0〉$ .

It is observed that the position vector and the principal unit normal vector are not equal.

Therefore, the given statement is false.

Answer 42

f.

To determine

Whether the statement “The position vector and the principal unit normal are always parallel on a smooth curve.” is true or not.

Answer 43

f.

Expert Solution

Answer to Problem 1RE

The given statement is false.

Answer 44

f.

Expert Solution

Answer 45

f.

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Formula used:

Suppose r is a smooth parameterized curve and s is the arc length.

The unit tangent vector T is $r′|r′|$ .

The principal unit normal vector is $N(t)=T′(t)|T′(t)|$ .

Counter example:

Consider $r(t)=〈4sint,4cost,10t〉$ .

Differentiate $r(t)$ to compute $r′(t)$ .

$r′(t)=〈(4sint)′,(4cost)′,(10t)′〉=〈4cost,−4sint,10〉$

Use magnitude formula to obtain the value of $|r′(t)|$ .

$|r′(t)|=|〈4cost,−4sint,10〉|=(4cost)2+(−4sint)2+102=16cos2t+16sin2t+100=16(cos2t+sin2t)+100$

On further simplification,

$|r′(t)|=16(1)+100=116=229$

Use unit tangent formula to compute $T(t)$ .

$T(t)=〈4cost,−4sint,10〉229=129〈4cost2,−4sint2,102〉=129〈2cost,−2sint,5〉$

Thus, the unit tangent vector $T(t)$ is $129〈2cost,−2sint,5〉$ .

Differentiate $T(t)$ to compute $T′(t)$ .

$T′(t)=129〈(2cost)′,(−2sint)′,5′〉=129〈−2sint,−2cost,0〉$

Use magnitude formula to obtain the value of $|T′(t)|$ .

$|T′(t)|=|129〈−2sint,−2cost,0〉|=129(−2sint)2+(−2cost)2+02=1294sin2t+4cos2t=1294(sin2t+cos2t)$

On further simplification,

$|T′(t)|=1294(1)=229$

Use principal unit normal formula to compute the value of $N(t)$ .

$N(t)=129〈−2sint,−2cost,0〉229=129(292)〈−2sint,−2cost,0〉=12〈−2sint,−2cost,0〉=〈−sint2,−cost2,02〉$

$=〈−sint,−cost,0〉$

Thus, the principal unit normal vector $N(t)$ for the curve $r(t)$ is $〈−sint,−cost,0〉$ .

It is observed that the position vector and the principal unit normal vector are not equal.

Therefore, the given statement is false.

Answer 48

Ch. 14.1 - Restrict the domain o f the vector function in...Ch. 14.1 - Explain why the curve in Example 5 lies on the...Ch. 14.1 - How many independent variables does the function...Ch. 14.1 - How many dependent scalar variables does the...Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - How do you evaluate limtar(t), where r(t) = f(t),...Ch. 14.1 - How do you determine whether r(t) = f(t) i + g(t)...Ch. 14.1 - Find a function r(t) for the line passing through...Ch. 14.1 - Find a function r(t) whose graph is a circle of...

Whether the statement “If r ( t ) = 〈 cos t , e t , t 〉 + C and r ( 0 ) = 〈 0 , 0 , 0 〉 , then C = 〈 0 , 0 , 0 〉 ” is true or not.

Concept explainers

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Answer to Problem 1RE

Explanation of Solution

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Answer to Problem 1RE

Explanation of Solution

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Answer to Problem 1RE

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Answer to Problem 1RE

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Answer to Problem 1RE

Explanation of Solution

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Answer to Problem 1RE

Explanation of Solution

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Whether the statement “If r ( t ) = 〈 cos t , e t , t 〉 + C and r ( 0 ) = 〈 0 , 0 , 0 〉 , then C = 〈 0 , 0 , 0 〉 ” is true or not.

Concept explainers

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Answer to Problem 1RE

Explanation of Solution

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Explanation of Solution

Answer to Problem 1RE

Explanation of Solution

Answer to Problem 1RE

Explanation of Solution

Answer to Problem 1RE

Explanation of Solution

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Answer to Problem 1RE

Explanation of Solution

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Chapter 14 Solutions

Additional Math Textbook Solutions

Whether the statement “The curvature of a circle of radius 5 is $κ=15$ ” is true or not.