Modeling the Dynamics of Life: Calculus and Probability for Life Scientists
Modeling the Dynamics of Life: Calculus and Probability for Life Scientists
3rd Edition
ISBN: 9780840064189
Author: Frederick R. Adler
Publisher: Cengage Learning
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Chapter 2.9, Problem 38E
To determine

To find:The slopes of tangents to the given relations at given points.

Expert Solution & Answer
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Answer to Problem 38E

.

The slopes of tangents at given points are given in table below:

    xslope
    -0.5- 1
    +0.5- 1
    -1+8
    +1- 8
    0undefined

The graph of the curve is shown in figure.

Explanation of Solution

Given information:

The given implicit relation is y2+4xy+4x4=0 , the slopes of tangents at points x=− 0.5 , x=0.5 x=−1 x=1 is required. The slope at of tangent line at x=0 is also to be discussed.

Method/Formulaused:

The slope of tangent line ( m ) at point (a,b) to the curve y=f(x) is given by m=[dydx](a,b) .

Calculations:

The given relation that represents a curve is

  y2+4xy+4x4=0   ... (1)

Differentiating with respect to ‘ x’

  2ydydx+4(xdydx+y)+16x3=0  ... (2)

  ⇒(2y+4x)dydx=−(16x3+4y)

  ⇒dydx=−(16x3+4y)(2y+4x)=−(y+8x3)(y+2x)  ... (3)

Now, from relation (1) y2+4xy+4x4=0

  ⇒y=−4x±16x2−16x42=−4x±4x1−x22=−2x±2x1−x2 .

Now, for x=− 0.5 , we have y=2x(−1±1−x2)=2(−0.5)(−1±1− ( −0.5 )2)

  =−1(−1±0.75)=(1±0.75)=0.1340,1.8660 .

Hence, the slope of tangent at x=− 0.5 are

  [dydx]x  =  −  0.5=−(y+8x3)(y+2x)=−(0.1340+8 ( −0.5 )3)(0.1340+2( −0.5)),−(1.8660+8 ( −0.5 )3)(1.8660+2( −0.5))

  [dydx]x  =   −  0.5=−1,−1   ... (4)

Now, for x=0.5 , we have y=2x(−1±1−x2)=2(0.5)(−1±1− ( 0.5 )2)=(−1±0.75)=−0.1340,−1.8660

Hence, the slope of tangent at x= 0.5 are

  [dydx]x =  0.5=−(y+8x3)(y+2x)=−(−0.1340+8 ( 0.5 )3)(−0.1340+2( 0.5)),−(−1.8660+8 ( 0.5 )3)(−1.8660+2( 0.5))

  [dydx]x =  0.5=−(−0.1340+1)(−0.1340+1),−(−1.8660+1)(−1.8660+1)=−1,−1  ... (5)

Now, for x=−1 , we have

  y=2x(−1±1−x2)=2(−1)(−1±1− ( −1 )2)=−2(−1±0)=2,2

Hence, the slope of tangent at x= −1 are

  [dydx]x =  −1=−(y+8x3)(y+2x)=−(2+8 ( −1 )3)(2+2( −1)),−(2+8 ( −1 )3)(2+2( −1))=∞,∞   ... (6)

Now, for x=1 , we have

  y=2x(−1±1−x2)=2(1)(−1±1− ( 1 )2)=2(−1±0)=−2,−2

Hence, the slope of tangent at x= 1 are

  [dydx]x =  1=−(y+8x3)(y+2x)=−(−2+8 ( 1 )3)(−2+2(1)),−(−2+8 ( 1 )3)(−2+2(1))=−∞,−∞   ... (7)

Now, for x=0 , we have

  y=2x(−1±1−x2)=0

Hence, the slope of tangent at x= 0 are

  [dydx]x =  0.5=−(y+8x3)(y+2x)=00 ,

that is undefined.

The graph of the curve and tangents at given points are shown in figure here.

  Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.9, Problem 38E

Conclusion:

The slopes of tangents at given points are given in table below:

    xslope
    -0.5- 1
    +0.5- 1
    -1+8
    +1- 8
    0undefined

The graph of the curve is shown in figure.

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Chapter 2 Solutions

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists

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Prob. 27ECh. 2.3 - Prob. 28ECh. 2.3 - Prob. 29ECh. 2.3 - Prob. 30ECh. 2.3 - Prob. 31ECh. 2.3 - Prob. 32ECh. 2.3 - Prob. 33ECh. 2.3 - Prob. 34ECh. 2.3 - Prob. 35ECh. 2.3 - Prob. 36ECh. 2.3 - Prob. 37ECh. 2.3 - Prob. 38ECh. 2.3 - Prob. 39ECh. 2.3 - Prob. 40ECh. 2.3 - Prob. 41ECh. 2.3 - Prob. 42ECh. 2.3 - Prob. 43ECh. 2.3 - Prob. 44ECh. 2.3 - Prob. 45ECh. 2.3 - Prob. 46ECh. 2.3 - Prob. 47ECh. 2.4 - Prob. 1ECh. 2.4 - Prob. 2ECh. 2.4 - Prob. 3ECh. 2.4 - Prob. 4ECh. 2.4 - Prob. 5ECh. 2.4 - Prob. 6ECh. 2.4 - Prob. 7ECh. 2.4 - Prob. 8ECh. 2.4 - Prob. 9ECh. 2.4 - Prob. 10ECh. 2.4 - Prob. 11ECh. 2.4 - For each of the following quadratic functions,...Ch. 2.4 - Prob. 13ECh. 2.4 - Prob. 14ECh. 2.4 - Prob. 15ECh. 2.4 - Prob. 16ECh. 2.4 - Prob. 17ECh. 2.4 - Prob. 18ECh. 2.4 - Prob. 19ECh. 2.4 - Prob. 20ECh. 2.4 - Prob. 21ECh. 2.4 - Prob. 22ECh. 2.4 - Prob. 23ECh. 2.4 - Prob. 24ECh. 2.4 - Prob. 25ECh. 2.4 - Prob. 26ECh. 2.4 - Prob. 27ECh. 2.4 - Prob. 28ECh. 2.4 - Prob. 29ECh. 2.4 - Prob. 30ECh. 2.4 - 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