The derivative of given function, to show change in the value of function with geometrical diagram, to give geometrical interpretation geometrically, and tell if unit makes sense.

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Answer 1

Question

Chapter 2.5, Problem 30E

To determine

To find:The derivative of given function, to show change in the value of function with geometrical diagram, to give geometrical interpretation geometrically, and tell if unit makes sense.

Expert Solution & Answer

Answer to Problem 30E

The derivative of volume with respect to radius is $dVdr=4πr2$ .

Explanation of Solution

Given information:

The volume of a sphere as a function of radius r is given by $V(r)=4π3r3$ . We have to find derivative of volume with respect to radius and show change in volume $△V=V(r+△r)−V(r)$ with geometrical diagram, to give geometrical interpretation of the derivative, and to tell if units make sense.

Method/Formulaused:

The derivative of the function $y=Cxn$ , C constant is $dydx=Cnxn−1$ .

Calculations:

The volumeof sphere of radiusris given by

$V(r)=4π3r3$ ... (1)

Differentiating equation (1) with respect to ‘ r’ , we have

$dVdr=4π3×3r2=4πr2$ ... (2)

The derivative of volume gives the change in volume of a sphere relative to the radius.

Let initially the radius of a sphere is r and changed to $r+△r$ , then change in volume is

$△V=V(r+△r)−V(r)$ ... (3)

$⇒△V=4π3(r+△r)3−4π3(r)3=4π3[(r+△r)3−(r)3]$

$⇒△V=4π3[r3+(△r)3+3r△r(r+△r)−r3]=4π3[(△r)3+3r2△r+3r(△r)2]$

Thus, $V(r+△r)−V(r)=4π3[(△r)3+3r2△r+3r(△r)2]$ .

Now, $V(r+△r)−V(r)△r=4π3[(△r)2+3r2+3r(△r)]$

Since $(△r)2$ is very small quantity so can be neglected, therefore,

Let we write $V(r+△r)−V(r)△r=△V△r$

$lim△r→0△V△r=lim△r→04π3[(△r)2+3r2+3r(△r)]$

$⇒lim△r→0△V△r=4π33r2=4πr2$

$⇒dVdr=4πr2$ $∵lim△r→0△V△r=dVdr$ .

The change in volume of sphere with radiusis geometrically shown in Figure-1 here,

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.5, Problem 30E , additional homework tip 1

Interpretation of derivative geometrically:

It is given that the volume of sphere is a function of its radius. The graph of Vasthe function r is shown in Figure-2.

We consider two points $P(r,V)$ and $Q(r+△r,V+△V)$ . Clearly line PQ is secant and it will become a tangent when $Q→P⇒△r→0$ .

Let $∠QPN=θ$ , then from Figure-2, $tanθ=QNPN=△V△r$ .

Now the secant PQ becomes tangent,then taking limit as $△r→0$ , we have $lim△r→0tanθ=lim△r→0△V△r$

$⇒tanθ=dVdr$ . But $tanθ$ is the slope of tangent to the curve at point P.

Thus, the derivative of V with respect to radius r represents the slope of the tangent line at a point of the curve,

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.5, Problem 30E , additional homework tip 2

The unit of $△V$ is the as that of volume V and the unit of $△r$ is same as that of radius r therefore the unit of the derivative is unit of volume divided by the unit of radius.

Conclusion:

The derivative of volume with respect to radius is $dVdr=4πr2$ .

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Answer 2

Question

Chapter 2.5, Problem 30E

To determine

To find:The derivative of given function, to show change in the value of function with geometrical diagram, to give geometrical interpretation geometrically, and tell if unit makes sense.

Expert Solution & Answer

Answer to Problem 30E

The derivative of volume with respect to radius is $dVdr=4πr2$ .

Explanation of Solution

Given information:

The volume of a sphere as a function of radius r is given by $V(r)=4π3r3$ . We have to find derivative of volume with respect to radius and show change in volume $△V=V(r+△r)−V(r)$ with geometrical diagram, to give geometrical interpretation of the derivative, and to tell if units make sense.

Method/Formulaused:

The derivative of the function $y=Cxn$ , C constant is $dydx=Cnxn−1$ .

Calculations:

The volumeof sphere of radiusris given by

$V(r)=4π3r3$ ... (1)

Differentiating equation (1) with respect to ‘ r’ , we have

$dVdr=4π3×3r2=4πr2$ ... (2)

The derivative of volume gives the change in volume of a sphere relative to the radius.

Let initially the radius of a sphere is r and changed to $r+△r$ , then change in volume is

$△V=V(r+△r)−V(r)$ ... (3)

$⇒△V=4π3(r+△r)3−4π3(r)3=4π3[(r+△r)3−(r)3]$

$⇒△V=4π3[r3+(△r)3+3r△r(r+△r)−r3]=4π3[(△r)3+3r2△r+3r(△r)2]$

Thus, $V(r+△r)−V(r)=4π3[(△r)3+3r2△r+3r(△r)2]$ .

Now, $V(r+△r)−V(r)△r=4π3[(△r)2+3r2+3r(△r)]$

Since $(△r)2$ is very small quantity so can be neglected, therefore,

Let we write $V(r+△r)−V(r)△r=△V△r$

$lim△r→0△V△r=lim△r→04π3[(△r)2+3r2+3r(△r)]$

$⇒lim△r→0△V△r=4π33r2=4πr2$

$⇒dVdr=4πr2$ $∵lim△r→0△V△r=dVdr$ .

The change in volume of sphere with radiusis geometrically shown in Figure-1 here,

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.5, Problem 30E , additional homework tip 1

Interpretation of derivative geometrically:

It is given that the volume of sphere is a function of its radius. The graph of Vasthe function r is shown in Figure-2.

We consider two points $P(r,V)$ and $Q(r+△r,V+△V)$ . Clearly line PQ is secant and it will become a tangent when $Q→P⇒△r→0$ .

Let $∠QPN=θ$ , then from Figure-2, $tanθ=QNPN=△V△r$ .

Now the secant PQ becomes tangent,then taking limit as $△r→0$ , we have $lim△r→0tanθ=lim△r→0△V△r$

$⇒tanθ=dVdr$ . But $tanθ$ is the slope of tangent to the curve at point P.

Thus, the derivative of V with respect to radius r represents the slope of the tangent line at a point of the curve,

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.5, Problem 30E , additional homework tip 2

The unit of $△V$ is the as that of volume V and the unit of $△r$ is same as that of radius r therefore the unit of the derivative is unit of volume divided by the unit of radius.

Conclusion:

The derivative of volume with respect to radius is $dVdr=4πr2$ .

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Answer 3

Question

Chapter 2.5, Problem 30E

To determine

To find:The derivative of given function, to show change in the value of function with geometrical diagram, to give geometrical interpretation geometrically, and tell if unit makes sense.

Expert Solution & Answer

Answer to Problem 30E

The derivative of volume with respect to radius is $dVdr=4πr2$ .

Explanation of Solution

Given information:

The volume of a sphere as a function of radius r is given by $V(r)=4π3r3$ . We have to find derivative of volume with respect to radius and show change in volume $△V=V(r+△r)−V(r)$ with geometrical diagram, to give geometrical interpretation of the derivative, and to tell if units make sense.

Method/Formulaused:

The derivative of the function $y=Cxn$ , C constant is $dydx=Cnxn−1$ .

Calculations:

The volumeof sphere of radiusris given by

$V(r)=4π3r3$ ... (1)

Differentiating equation (1) with respect to ‘ r’ , we have

$dVdr=4π3×3r2=4πr2$ ... (2)

The derivative of volume gives the change in volume of a sphere relative to the radius.

Let initially the radius of a sphere is r and changed to $r+△r$ , then change in volume is

$△V=V(r+△r)−V(r)$ ... (3)

$⇒△V=4π3(r+△r)3−4π3(r)3=4π3[(r+△r)3−(r)3]$

$⇒△V=4π3[r3+(△r)3+3r△r(r+△r)−r3]=4π3[(△r)3+3r2△r+3r(△r)2]$

Thus, $V(r+△r)−V(r)=4π3[(△r)3+3r2△r+3r(△r)2]$ .

Now, $V(r+△r)−V(r)△r=4π3[(△r)2+3r2+3r(△r)]$

Since $(△r)2$ is very small quantity so can be neglected, therefore,

Let we write $V(r+△r)−V(r)△r=△V△r$

$lim△r→0△V△r=lim△r→04π3[(△r)2+3r2+3r(△r)]$

$⇒lim△r→0△V△r=4π33r2=4πr2$

$⇒dVdr=4πr2$ $∵lim△r→0△V△r=dVdr$ .

The change in volume of sphere with radiusis geometrically shown in Figure-1 here,

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.5, Problem 30E , additional homework tip 1

Interpretation of derivative geometrically:

It is given that the volume of sphere is a function of its radius. The graph of Vasthe function r is shown in Figure-2.

We consider two points $P(r,V)$ and $Q(r+△r,V+△V)$ . Clearly line PQ is secant and it will become a tangent when $Q→P⇒△r→0$ .

Let $∠QPN=θ$ , then from Figure-2, $tanθ=QNPN=△V△r$ .

Now the secant PQ becomes tangent,then taking limit as $△r→0$ , we have $lim△r→0tanθ=lim△r→0△V△r$

$⇒tanθ=dVdr$ . But $tanθ$ is the slope of tangent to the curve at point P.

Thus, the derivative of V with respect to radius r represents the slope of the tangent line at a point of the curve,

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.5, Problem 30E , additional homework tip 2

The unit of $△V$ is the as that of volume V and the unit of $△r$ is same as that of radius r therefore the unit of the derivative is unit of volume divided by the unit of radius.

Conclusion:

The derivative of volume with respect to radius is $dVdr=4πr2$ .

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Answer 4

Question

Chapter 2.5, Problem 30E

To determine

To find:The derivative of given function, to show change in the value of function with geometrical diagram, to give geometrical interpretation geometrically, and tell if unit makes sense.

Expert Solution & Answer

Answer to Problem 30E

The derivative of volume with respect to radius is $dVdr=4πr2$ .

Explanation of Solution

Given information:

The volume of a sphere as a function of radius r is given by $V(r)=4π3r3$ . We have to find derivative of volume with respect to radius and show change in volume $△V=V(r+△r)−V(r)$ with geometrical diagram, to give geometrical interpretation of the derivative, and to tell if units make sense.

Method/Formulaused:

The derivative of the function $y=Cxn$ , C constant is $dydx=Cnxn−1$ .

Calculations:

The volumeof sphere of radiusris given by

$V(r)=4π3r3$ ... (1)

Differentiating equation (1) with respect to ‘ r’ , we have

$dVdr=4π3×3r2=4πr2$ ... (2)

The derivative of volume gives the change in volume of a sphere relative to the radius.

Let initially the radius of a sphere is r and changed to $r+△r$ , then change in volume is

$△V=V(r+△r)−V(r)$ ... (3)

$⇒△V=4π3(r+△r)3−4π3(r)3=4π3[(r+△r)3−(r)3]$

$⇒△V=4π3[r3+(△r)3+3r△r(r+△r)−r3]=4π3[(△r)3+3r2△r+3r(△r)2]$

Thus, $V(r+△r)−V(r)=4π3[(△r)3+3r2△r+3r(△r)2]$ .

Now, $V(r+△r)−V(r)△r=4π3[(△r)2+3r2+3r(△r)]$

Since $(△r)2$ is very small quantity so can be neglected, therefore,

Let we write $V(r+△r)−V(r)△r=△V△r$

$lim△r→0△V△r=lim△r→04π3[(△r)2+3r2+3r(△r)]$

$⇒lim△r→0△V△r=4π33r2=4πr2$

$⇒dVdr=4πr2$ $∵lim△r→0△V△r=dVdr$ .

The change in volume of sphere with radiusis geometrically shown in Figure-1 here,

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.5, Problem 30E , additional homework tip 1

Interpretation of derivative geometrically:

It is given that the volume of sphere is a function of its radius. The graph of Vasthe function r is shown in Figure-2.

We consider two points $P(r,V)$ and $Q(r+△r,V+△V)$ . Clearly line PQ is secant and it will become a tangent when $Q→P⇒△r→0$ .

Let $∠QPN=θ$ , then from Figure-2, $tanθ=QNPN=△V△r$ .

Now the secant PQ becomes tangent,then taking limit as $△r→0$ , we have $lim△r→0tanθ=lim△r→0△V△r$

$⇒tanθ=dVdr$ . But $tanθ$ is the slope of tangent to the curve at point P.

Thus, the derivative of V with respect to radius r represents the slope of the tangent line at a point of the curve,

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.5, Problem 30E , additional homework tip 2

The unit of $△V$ is the as that of volume V and the unit of $△r$ is same as that of radius r therefore the unit of the derivative is unit of volume divided by the unit of radius.

Conclusion:

The derivative of volume with respect to radius is $dVdr=4πr2$ .

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Answer 5

Chapter 2.5, Problem 30E

To determine

To find:The derivative of given function, to show change in the value of function with geometrical diagram, to give geometrical interpretation geometrically, and tell if unit makes sense.

Expert Solution & Answer

Answer to Problem 30E

The derivative of volume with respect to radius is $dVdr=4πr2$ .

Explanation of Solution

Given information:

The volume of a sphere as a function of radius r is given by $V(r)=4π3r3$ . We have to find derivative of volume with respect to radius and show change in volume $△V=V(r+△r)−V(r)$ with geometrical diagram, to give geometrical interpretation of the derivative, and to tell if units make sense.

Method/Formulaused:

The derivative of the function $y=Cxn$ , C constant is $dydx=Cnxn−1$ .

Calculations:

The volumeof sphere of radiusris given by

$V(r)=4π3r3$ ... (1)

Differentiating equation (1) with respect to ‘ r’ , we have

$dVdr=4π3×3r2=4πr2$ ... (2)

The derivative of volume gives the change in volume of a sphere relative to the radius.

Let initially the radius of a sphere is r and changed to $r+△r$ , then change in volume is

$△V=V(r+△r)−V(r)$ ... (3)

$⇒△V=4π3(r+△r)3−4π3(r)3=4π3[(r+△r)3−(r)3]$

$⇒△V=4π3[r3+(△r)3+3r△r(r+△r)−r3]=4π3[(△r)3+3r2△r+3r(△r)2]$

Thus, $V(r+△r)−V(r)=4π3[(△r)3+3r2△r+3r(△r)2]$ .

Now, $V(r+△r)−V(r)△r=4π3[(△r)2+3r2+3r(△r)]$

Since $(△r)2$ is very small quantity so can be neglected, therefore,

Let we write $V(r+△r)−V(r)△r=△V△r$

$lim△r→0△V△r=lim△r→04π3[(△r)2+3r2+3r(△r)]$

$⇒lim△r→0△V△r=4π33r2=4πr2$

$⇒dVdr=4πr2$ $∵lim△r→0△V△r=dVdr$ .

The change in volume of sphere with radiusis geometrically shown in Figure-1 here,

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.5, Problem 30E , additional homework tip 1

Interpretation of derivative geometrically:

It is given that the volume of sphere is a function of its radius. The graph of Vasthe function r is shown in Figure-2.

We consider two points $P(r,V)$ and $Q(r+△r,V+△V)$ . Clearly line PQ is secant and it will become a tangent when $Q→P⇒△r→0$ .

Let $∠QPN=θ$ , then from Figure-2, $tanθ=QNPN=△V△r$ .

Now the secant PQ becomes tangent,then taking limit as $△r→0$ , we have $lim△r→0tanθ=lim△r→0△V△r$

$⇒tanθ=dVdr$ . But $tanθ$ is the slope of tangent to the curve at point P.

Thus, the derivative of V with respect to radius r represents the slope of the tangent line at a point of the curve,

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.5, Problem 30E , additional homework tip 2

The unit of $△V$ is the as that of volume V and the unit of $△r$ is same as that of radius r therefore the unit of the derivative is unit of volume divided by the unit of radius.

Conclusion:

The derivative of volume with respect to radius is $dVdr=4πr2$ .

Answer 6

Chapter 2.5, Problem 30E

To determine

To find:The derivative of given function, to show change in the value of function with geometrical diagram, to give geometrical interpretation geometrically, and tell if unit makes sense.

Expert Solution & Answer

Answer to Problem 30E

The derivative of volume with respect to radius is $dVdr=4πr2$ .

Explanation of Solution

Given information:

The volume of a sphere as a function of radius r is given by $V(r)=4π3r3$ . We have to find derivative of volume with respect to radius and show change in volume $△V=V(r+△r)−V(r)$ with geometrical diagram, to give geometrical interpretation of the derivative, and to tell if units make sense.

Method/Formulaused:

The derivative of the function $y=Cxn$ , C constant is $dydx=Cnxn−1$ .

Calculations:

The volumeof sphere of radiusris given by

$V(r)=4π3r3$ ... (1)

Differentiating equation (1) with respect to ‘ r’ , we have

$dVdr=4π3×3r2=4πr2$ ... (2)

The derivative of volume gives the change in volume of a sphere relative to the radius.

Let initially the radius of a sphere is r and changed to $r+△r$ , then change in volume is

$△V=V(r+△r)−V(r)$ ... (3)

$⇒△V=4π3(r+△r)3−4π3(r)3=4π3[(r+△r)3−(r)3]$

$⇒△V=4π3[r3+(△r)3+3r△r(r+△r)−r3]=4π3[(△r)3+3r2△r+3r(△r)2]$

Thus, $V(r+△r)−V(r)=4π3[(△r)3+3r2△r+3r(△r)2]$ .

Now, $V(r+△r)−V(r)△r=4π3[(△r)2+3r2+3r(△r)]$

Since $(△r)2$ is very small quantity so can be neglected, therefore,

Let we write $V(r+△r)−V(r)△r=△V△r$

$lim△r→0△V△r=lim△r→04π3[(△r)2+3r2+3r(△r)]$

$⇒lim△r→0△V△r=4π33r2=4πr2$

$⇒dVdr=4πr2$ $∵lim△r→0△V△r=dVdr$ .

The change in volume of sphere with radiusis geometrically shown in Figure-1 here,

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.5, Problem 30E , additional homework tip 1

Interpretation of derivative geometrically:

It is given that the volume of sphere is a function of its radius. The graph of Vasthe function r is shown in Figure-2.

We consider two points $P(r,V)$ and $Q(r+△r,V+△V)$ . Clearly line PQ is secant and it will become a tangent when $Q→P⇒△r→0$ .

Let $∠QPN=θ$ , then from Figure-2, $tanθ=QNPN=△V△r$ .

Now the secant PQ becomes tangent,then taking limit as $△r→0$ , we have $lim△r→0tanθ=lim△r→0△V△r$

$⇒tanθ=dVdr$ . But $tanθ$ is the slope of tangent to the curve at point P.

Thus, the derivative of V with respect to radius r represents the slope of the tangent line at a point of the curve,

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.5, Problem 30E , additional homework tip 2

The unit of $△V$ is the as that of volume V and the unit of $△r$ is same as that of radius r therefore the unit of the derivative is unit of volume divided by the unit of radius.

Conclusion:

The derivative of volume with respect to radius is $dVdr=4πr2$ .

Answer 7

Chapter 2.5, Problem 30E

To determine

To find:The derivative of given function, to show change in the value of function with geometrical diagram, to give geometrical interpretation geometrically, and tell if unit makes sense.

Answer 8

Expert Solution & Answer

Answer to Problem 30E

The derivative of volume with respect to radius is $dVdr=4πr2$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The volume of a sphere as a function of radius r is given by $V(r)=4π3r3$ . We have to find derivative of volume with respect to radius and show change in volume $△V=V(r+△r)−V(r)$ with geometrical diagram, to give geometrical interpretation of the derivative, and to tell if units make sense.

Method/Formulaused:

The derivative of the function $y=Cxn$ , C constant is $dydx=Cnxn−1$ .

Calculations:

The volumeof sphere of radiusris given by

$V(r)=4π3r3$ ... (1)

Differentiating equation (1) with respect to ‘ r’ , we have

$dVdr=4π3×3r2=4πr2$ ... (2)

The derivative of volume gives the change in volume of a sphere relative to the radius.

Let initially the radius of a sphere is r and changed to $r+△r$ , then change in volume is

$△V=V(r+△r)−V(r)$ ... (3)

$⇒△V=4π3(r+△r)3−4π3(r)3=4π3[(r+△r)3−(r)3]$

$⇒△V=4π3[r3+(△r)3+3r△r(r+△r)−r3]=4π3[(△r)3+3r2△r+3r(△r)2]$

Thus, $V(r+△r)−V(r)=4π3[(△r)3+3r2△r+3r(△r)2]$ .

Now, $V(r+△r)−V(r)△r=4π3[(△r)2+3r2+3r(△r)]$

Since $(△r)2$ is very small quantity so can be neglected, therefore,

Let we write $V(r+△r)−V(r)△r=△V△r$

$lim△r→0△V△r=lim△r→04π3[(△r)2+3r2+3r(△r)]$

$⇒lim△r→0△V△r=4π33r2=4πr2$

$⇒dVdr=4πr2$ $∵lim△r→0△V△r=dVdr$ .

The change in volume of sphere with radiusis geometrically shown in Figure-1 here,

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.5, Problem 30E , additional homework tip 1

Interpretation of derivative geometrically:

It is given that the volume of sphere is a function of its radius. The graph of Vasthe function r is shown in Figure-2.

We consider two points $P(r,V)$ and $Q(r+△r,V+△V)$ . Clearly line PQ is secant and it will become a tangent when $Q→P⇒△r→0$ .

Let $∠QPN=θ$ , then from Figure-2, $tanθ=QNPN=△V△r$ .

Now the secant PQ becomes tangent,then taking limit as $△r→0$ , we have $lim△r→0tanθ=lim△r→0△V△r$

$⇒tanθ=dVdr$ . But $tanθ$ is the slope of tangent to the curve at point P.

Thus, the derivative of V with respect to radius r represents the slope of the tangent line at a point of the curve,

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.5, Problem 30E , additional homework tip 2

The unit of $△V$ is the as that of volume V and the unit of $△r$ is same as that of radius r therefore the unit of the derivative is unit of volume divided by the unit of radius.

Conclusion:

The derivative of volume with respect to radius is $dVdr=4πr2$ .