Chapter 4.4, Problem 22E

To determine

To Find:The exact coordinates of the function.

Answer to Problem 22E

  $(-0.25,0.3605)$

Explanation of Solution

Given:Use a graph to estimate the coordinates of the leftmost point on the curve $x=t^4-t^2,y=t+\ln t$ then uses the calculus to find the exact coordinates.

Given: $x=t^4-t^2,y=t+\ln t$

The graph of the function is

  

The leftmost point is $x=-0.2$

Then

It means that the slope of tangent is at the point $\infty$

Then

  $\begin{array}{l}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\infty \\ \frac{dx}{dt}=0\\ \frac{d(t^4-t^2)}{dt}=4t^3-2t=2t(2t^2-1)=0\end{array}$

Factoring we get

  $\begin{array}{l}2t(2t^2-1)=0\\ 2t=0\\ t=0\\ 2t^2-1=0\\ t=\pm \frac{1}{\sqrt{2}}\end{array}$

Now

  $\begin{array}{l}y=t+\ln t\\ t=\frac{1}{\sqrt{2}},t>0\end{array}$

Hence

The coordinates are

  $\begin{array}{l}x=t^4-t^2\\ ={\left(\frac{1}{\sqrt{2}}\right)}^4-{\left(\frac{1}{\sqrt{2}}\right)}^2\\ =-\frac{1}{4}\\ y=t+\ln t\\ y=\frac{1}{\sqrt{2}}+\ln \frac{1}{\sqrt{2}}\\ y=0.3605\\ coordinate:(-0.25,0.3605)\end{array}$