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 2.4, Problem 39E

a.

To determine

To graph: The secant line to the given function at base point t=1 with △t=1 .

a.

Expert Solution
Check Mark

Explanation of Solution

Given:

The given function is y(t)=10t2.5 . Base point is t=1 and secant joins the base point t=1 and the point with △t=1 .

Method used:

The function and the secant line is plotted using graphic calculator.

Calculations and Graph:

  Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.4, Problem 39E , additional homework tip 1

The given function is y(t)=10t2.5 .

The coordinates of base point say, P are

  (t,10t2.5)≡(1,10⋅(1)2.5)=(1,10) .

The point at distance △t=1 from base point is

  Q(t+△t,10(t+△t)2.5)≡(1+1,10⋅(1+1)2.5)

  ⇒Q=(2,10⋅252)=(2,402)

The graph of the function and the secant is shown in Figure (a)

b.

To determine

To graph: The secant line to the given function at base point t=1 with △t=−1 .

b.

Expert Solution
Check Mark

Explanation of Solution

Given:

The given function is y(t)=10t2.5 . Base point is t=1 and secant joins the base point t=1 and the point with △t=−1 .

Method used:

The function and the secant line is plotted using graphic calculator.

Calculations and Graph:

  Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.4, Problem 39E , additional homework tip 2

The given function is y(t)=10t2.5 .

The coordinates of base point say, P are (t,10t2.5)≡(1,10⋅(1)2.5)=(1,10) .

The point at distance △t=1 from base point is Q(t+△t,10(t+△t)2.5)

  ⇒Q≡(1−1,10⋅(1−1)2.5)

  ⇒Q≡(1−1,10⋅(1−1)2.5)=(0,0) , that is the origin.

The graph of the function and the secant is shown in Figure (b).

c.

To determine

To graph: The secant line to the given function at base point t=1 with △t=0.1 .

c.

Expert Solution
Check Mark

Explanation of Solution

Given:

The given function is y(t)=10t2.5 . Base point is t=1 and secant joins the base point t=1 and the point with △t=0.1 .

Method used:

The function and the secant line is plotted using graphic calculator.

Calculations and Graph:

  Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.4, Problem 39E , additional homework tip 3

The given function is y(t)=10t2.5 .

The coordinates of base point say, P are (t,10t2.5)≡(1,10⋅(1)2.5)=(1,10) .

The point at distance △t=1 from base point is Q(t+△t,10(t+△t)2.5)

  ⇒Q≡(1+0.1,10⋅(1+0.1)2.5)

  ⇒Q=(1.1,10(1.1)5/2)=(1.1,12.69)

The graph of the function and the secant is shown in Figure (c). We see the secant line tends to tangent line at point P .

d.

To determine

To graph: The secant line to the given function at base point t=1 with △t=0.1 .

d.

Expert Solution
Check Mark

Explanation of Solution

Given:

The given function is y(t)=10t2.5 . Base point is t=1 and secant joins the base point t=1 and the point with △t=−0.1 .

Method used:

The function and the secant line is plotted using graphic calculator.

Calculations and Graph:

  Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.4, Problem 39E , additional homework tip 4

The given function is y(t)=10t2.5 .

The coordinates of base point say, P are (t,10t2.5)≡(1,10⋅(1)2.5)=(1,10) .

The point at distance △t=1 from base point is Q(t+△t,10(t+△t)2.5)

  ⇒Q≡(1−0.1,10⋅(1−0.1)2.5)

  ⇒Q=(0.9,10(0.9)5/2)=(0.9,7.68)

The graph of the function and the secant is shown in Figure (d).

We see the secant line tends to tangent line at point P .

e.

To determine

To graph: The secant line to the given function at base point t=1 with △t→0 , let △t=0.01 .

e.

Expert Solution
Check Mark

Explanation of Solution

Given:

The given function is y(t)=10t2.5 . Base point is t=1 and secant joins the base point t=1 and the point with assumed △t=0.01 .

Method used:

The function and the secant line is plotted using graphic calculator.

Calculations and Graph:

  Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.4, Problem 39E , additional homework tip 5

The given function is y(t)=10t2.5 .

The coordinates of base point say, P are (t,10t2.5)≡(1,10⋅(1)2.5)=(1,10) .

The point at distance △t=1 from base point is Q(t+△t,10(t+△t)2.5)

  ⇒Q≡(1+0.01,10⋅(1+0.01)2.5)

  ⇒Q=(1.01,10(1.01)5/2)=(1.01,10.25)

The graph of the function and the secant is shown in Figure (e).

We observe that points P(1,10) and Q(1.01,10.25) almost coincide, therefore the secant line and tangent line at point P(1,10) coincide. That we say, when △t→0 the secant line becomes a tangent line at base point. The tangent line at point P(Q→P) in this case looks right.

Here, the slope of the tangent by definition is m=tan∠PLM=PMLM=10.4=25 .

And from the formula as the tangent line is the limiting case of secant line m=y2−y1x2−x1=10.25−101.01−1.00=0.250.01=251=25 .

Thus, the slope of the tangent line is 25.

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

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