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a. For what values of r does the function y = erx satisfy the following equation? (Enter your...

Question:

a. For what values of r does the function y = erx satisfy the following equation? (Enter your answers from smallest to largest.)

{eq}y" + 8y' + 15y = 0 {/eq}

r =

r =

b. Find the absolute maximum value of the function.

{eq}f(x) = 3(x - e^x) {/eq}

maximum value =

c. Find an equation of the tangent line to the following curve at the given point.

{eq}y = e^{9x}cos(\pi x) (0,1) {/eq}

Applications of Calculus


This question is comprised of three parts. In part a. we substitute a general form exponential function into a second-order, ordinary differential equation to come up with the exact form for the exponential function. In part b. we find the maximum value of a function using its first derivative and the second derivative test. In part c. we use the first derivative to find the equation of the tangent to a function at a given point.


Answer and Explanation:


a. So we have the function

{eq}y=e^{rx} \implies y'=re^{rx} \; and \; y''=r^2 e^{rx} \qquad (1) {/eq}

using chain rule for derivatives from Calculus.

Then substituting the values from (1) into the given ordinary differential equation gives us:

{eq}r^2e^{rx}+8re^{rx}+15e^{rx}=0 \implies e^{rx}(r^2+8r+15)=0 \qquad (2) {/eq}

From (2) above, owing to the positive nature of the exponential function we get that {eq}r^2+8r+15=0 \implies (r+3)(r+5)=0 \implies r = -5, \; r = -3. {/eq}

Hence our answers are {eq}r = -5 \; and \; r = -3. {/eq}


b. To find the absolute maximum of the given function we try to find its critical points. To that end we calculate its first derivative as follows:

{eq}f'(x)=3-3e^x \qquad (3) {/eq}

Since f'(x) exists for all values of x we find critical points for f(x) by solving f'(x) = 0 from (3) above to get

{eq}3-3e^x=0 \implies 3 = 3e^x \implies e^x=1 \implies x=0. {/eq}

To ensure that x = 0 is a point of maximum for f(x) we calculate the second derivative of f(x) to arrive at

{eq}f''(x)=-3e^x \qquad (4) {/eq}

From (4) above, {eq}f''(0)=-3e^0=-3<0. {/eq}

Hence from the second derivative test, x = 0 is a point of maximum for f(x).

The absolute maximum value will then be {eq}f(0)=3(0-e^0)=3(-1)=-3. {/eq}


c. To find the equation of the tangent line to the given curve at point (x, y) = (0, 1) we need to find the slope m of the tangent line at that point. To that end we use Calculus and calculate the first derivative of the function and evaluate it at (x, y) = (0, 1) to get the slope of the tangent line at that point.

So the first derivative, using the product and chain rule from Calculus is

{eq}\displaystyle y'=\frac {d}{dx} e^{9x} \cos ( \pi x ) + e^{9x} \frac {d}{dx} \cos ( \pi x ) = 9e^{9x} \cos ( \pi x ) - \pi e^{9x} \sin ( \pi x) \qquad (5) {/eq}

From (5), the slope of the tangent line at (0, 1) will be given by

{eq}m=y'(0)=9e^{9(0)} \cos ( \pi (0) ) - \pi e^{9(0)} \sin ( \pi (0)) = 9 - 0 = 9. {/eq}

Using the point-slope form, the equation of the tangent line having slope {eq}m=9 {/eq} and passing through the point {eq}(x_1,y_1)=(0,1) {/eq} will be given by

{eq}y-y_1=m(x-x_1) \implies y-1=9(x-0) \implies y=9x+1. {/eq}

The equation of the tangent line at (0, 1) is {eq}y=9x+1. {/eq}


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Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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