Chapter 11.1, Problem 25E

(a)

To determine

To Find:The pointat which the tangent line to the curve in horizontal

Answer to Problem 25E

Theposition of the tangent are $\left(0.845,-3.079\right)and\left(3.155,3.079\right)$ .

Explanation of Solution

Given Information:

  $x=2-t$  ........ (1)

  $y=t^3-4t$  ........ (2)

Calculation:

Differentiating equation (1) and (2) with respect to $t$

  $\begin{array}{l}x=2-t\\ \frac{dx}{dt}=-1\end{array}$  ........ (3)

  $\begin{array}{l}y=t^3-4t\\ \frac{dy}{dt}=3t^2-4\end{array}$  ........ (4)

Now tangent of the curve is

  $\begin{array}{c}\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}\\ =4-3t^2\end{array}$

When

  $\begin{array}{c}4-3t^2=0\\ t=\pm \frac{2}{\sqrt{3}}\end{array}$

Substitute $\pm \frac{2}{\sqrt{3}}$ for $t$ in equation (1) and (2)

  $\begin{array}{c}\left(x,y\right)=\left(2\pm \frac{2}{\sqrt{3}},\pm {\left(\frac{2}{\sqrt{3}}\right)}^3\pm 4\left(\frac{2}{\sqrt{3}}\right)\right)\\ \simeq \left(0.845,-3.079\right)and\left(3.155,3.079\right)\end{array}$

Conclusion:

Hence, the position of the tangent are $\left(0.845,-3.079\right)and\left(3.155,3.079\right)$ .

(b)

To determine

To Find: The pointat which the tangent line to the curve in vertical.

Answer to Problem 25E

The vertical tangent doesn’t exist.

Explanation of Solution

Given Information:

  $x=2-t$  ........ (1)

  $y=t^3-4t$  ........ (2)

Calculation:

Differentiating equation (1) and (2) with respect to $t$

  $\frac{dx}{dt}=-1$  ........ (3)

  $\frac{dy}{dt}=3t-4$  ........ (4)

Now tangent of the curve is

  $\begin{array}{c}\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}\\ =\frac{3t-4}{-1}\\ =4-3t\end{array}$

The tangent line will be vertical when the equation for the tangent line $\left(\raisebox{1ex}{$\frac{dy}{dt}$}\!\left/ \!\raisebox{-1ex}{$\frac{dx}{dt}$}\right.\right)$ is undefined. This will occur when the denominator $\left(\frac{dx}{dt}\right)$ equal to $0$ . $-1$ Does not equal to $0$ .

Conclusion:

Hence, the vertical tangent doesn’t exist.