Parametric Functions: Use a graph to estimate the coordinates of the highest point and the...

Question:

Parametric Functions:

Use a graph to estimate the coordinates of the highest point and the leftmost point on the curve:

{eq}x = 6te^t , y = te^{-t}. {/eq}

Then find the exact coordinates.

Highest point:

Leftmost point:

The curve has one horizontal asymptote and vertical asymptote. Fine them.

Horizontal asymptote: y =

Vertical symptote: x =

Parametric Functions


Given a parametrically-defined function we find left most and highest point on the graph of this function. First we use a graph to estimate these points and then use calculus to find these points. This is followed by using calculus to find the horizontal and vertical asymptotes of this function. The horizontal and vertical asymptotes for a function occur at points where the first derivative is zero and undefined, respectively.


Answer and Explanation:


Using a graph we estimate the left-most point to be approximately at x = -3 and the highest point to be approximately at y = 0.3.

To find these points analytically we proceed as follows:


For the left most point we take the first derivative of x(t) to get

{eq}x'(t)=6te^t+6e^t = 6e^t(t+1) \qquad (1) {/eq}

Since x'(t) exists for all values of t we find critical points of x(t) by solving x'(t) = 0 to get t = -1. To ensure this a point of minimum which will give us the minimum x-value and therefore the left most point we calculate the second derivative of x(t) to get

{eq}x''(t)=6te^t+12e^t \qquad (2) {/eq}

From (2) {eq}x''(-1)=6e^{-1} > 0. {/eq} Hence from the second derivative test, t = -1 is a point of minimum for x(t) and so the left most point will be

{eq}(x(-1),y(-1))=(6(-1)e^{-1}, (-1)e^{1})=(-6/e, -e) {/eq}


For the highest point we take the first derivative of y(t) to get

{eq}y'(t)=-te^{-t}+e^{-t}=(-t+1)e^{-t} \qquad (3) {/eq}

Since y'(t) exists for all values of t we find critical points of y(t) by solving y'(t) = 0 to get t = 1. To ensure this a point of maximum which will give us the maximum y-value and therefore the highest point we calculate the second derivative of y(t) to get

{eq}y''(t)=(t-1)e^{-t}-e^{-t} = (t-2)e^{-t} \qquad (4) {/eq}

From (4) {eq}y''(1)=-e^{-1}<0. {/eq} Hence from the second derivative test, t = 1 is a point of maximum for y(t) and so the highest point will be

{eq}(x(1),y(1))=(6(1)e^{1}, (1)e^{-1})=(6e,1/e). {/eq}


To find the horizontal asymptote we look for where the graph has a horizontal tangent with slope zero. This would be a point on the graph where the derivative of y with respect to t will be zero. So we look for a solution to y'(t) = 0. We have already solved this equation above to get t = 1 and so the horizontal tangent will be at the highest point which is {eq}(x(1),y(1))=(6(1)e^{1}, (1)e^{-1})=(6e,1/e). {/eq}

Hence the horizontal asymptote will be {eq}y=1/e. {/eq}


To find the vertical asymptote we look for where the graph has a vertical tangent with slope undefined. This would be a point on the graph where the derivative of x with respect to t will be zero. So we look for a solution to x'(t) = 0. We have already solved this equation above to get t = -1 and so the horizontal tangent will be at the left most point which is {eq}(x(-1),y(-1))=(6(-1)e^{-1}, (-1)e^{1})=(-6/e, -e) {/eq}

Hence the vertical asymptote will be {eq}x=-6/e. {/eq}


Learn more about this topic:

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Graphs of Parametric Equations

from Precalculus: High School

Chapter 24 / Lesson 5
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