Chapter 3.4, Problem 85E

a.

To determine

Explain that the given curve has two tangent lines at the given point.

Answer to Problem 85E

The equation of tangent lines are $y=-\sqrt{3}x+3\sqrt{3}$ and $y=\sqrt{3}x-3\sqrt{3}$ .

Explanation of Solution

Given:

The given equations are $x=t^2$ and $y=t^3-3t$ . The given point is $\left(3,0\right)$ .

Calculation:

Find the slope.

Apply formula $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ .

  $\frac{dy}{dt}=\frac{d}{dt}\left(t^3-3t\right)$

Apply differencerule $\left(f-g\right)\text{'}=f\text{'}-g\text{'}$ .

  $\Rightarrow \frac{dy}{dt}=\frac{d}{dt}\left(t^3\right)-\frac{d}{dt}\left(3t\right)$

Use derivative rule $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

  $\begin{array}{l}\Rightarrow \frac{dy}{dt}=3t^2-3\\ \Rightarrow \frac{dy}{dt}=3\left(t^2-1\right)\end{array}$

  $\frac{dx}{dt}=\frac{d}{dt}\left(t^2\right)$

Use derivative rule $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

  $\Rightarrow \frac{dx}{dt}=2t$

  $\frac{dy}{dx}\left(\text{slope}\right)=\frac{3\left(t^2-1\right)}{2t}$

Find the value of the parameter $t$ at the point $\left(3,0\right)$ .

At $x=3$

  $\begin{array}{l}{t}^2=3\\ \Rightarrow t=\pm \sqrt{3}\end{array}$

At $y=0$

  $\begin{array}{l}{t}^3-3t=0\\ \Rightarrow t\left(t^2-3\right)=0\\ \Rightarrow t=0\\ \text{and}\\ \Rightarrow t^2-3=0\\ \Rightarrow t^2=3\\ \Rightarrow t=\pm \sqrt{3}\end{array}$

Now,

  $t=\pm \sqrt{3}$ , because it satisfies both the equations.

Slope at $t=-\sqrt{3}$ .

  $\begin{array}{l}\frac{dy}{dx}\left(\text{slope}\right)=\frac{3\left(t^2-1\right)}{2t}\\ =\frac{3\left[{\left(-\sqrt{3}\right)}^2-1\right]}{2\left(-\sqrt{3}\right)}\\ =\frac{3\left[3-1\right]}{2\left(-\sqrt{3}\right)}\\ =-\frac{3}{\sqrt{3}}\\ =-\frac{3}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ =-\sqrt{3}\end{array}$

Slope at $t=\sqrt{3}$ .

  $\begin{array}{l}\frac{dy}{dx}\left(\text{slope}\right)=\frac{3\left(t^2-1\right)}{2t}\\ =\frac{3\left[{\left(\sqrt{3}\right)}^2-1\right]}{2\left(\sqrt{3}\right)}\\ =\frac{3\left[3-1\right]}{2\left(\sqrt{3}\right)}\\ =\frac{3}{\sqrt{3}}\\ =\frac{3}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ =\sqrt{3}\end{array}$

There are two different tangents at the given point, because there is two different slope values.

Now use point-slope form for the tangent equations.

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y_1=0,x_1=3,m=-\sqrt{3}\\ y-0=-\sqrt{3}\left(x-3\right)\\ \Rightarrow y=-\sqrt{3}x+3\sqrt{3}\\ \text{and}\\ y_1=0,x_1=3,m=\sqrt{3}\\ y-0=\sqrt{3}\left(x-3\right)\\ \Rightarrow y=\sqrt{3}x-3\sqrt{3}\end{array}$

Hence the equation of tangent lines are $y=-\sqrt{3}x+3\sqrt{3}$ and $y=\sqrt{3}x-3\sqrt{3}$ .

b.

To determine

Find the horizontal and vertical tangent line points on the curve.

Answer to Problem 85E

The tangents are horizontal at the points $\left(2,2\right)$ and $\left(2,-2\right)$ . The tangent is vertical at the point $\left(0,0\right)$ .

Explanation of Solution

Given:

The given equations are $x=t^2$ and $y=t^3-3t$ . The given point is $\left(3,0\right)$ .

Calculation:

Find the slope.

Apply formula $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ .

  $\frac{dy}{dt}=\frac{d}{dt}\left(t^3-3t\right)$

Apply difference rule $\left(f-g\right)\text{'}=f\text{'}-g\text{'}$ .

  $\Rightarrow \frac{dy}{dt}=\frac{d}{dt}\left(t^3\right)-\frac{d}{dt}\left(3t\right)$

Use derivative rule $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

  $\begin{array}{l}\Rightarrow \frac{dy}{dt}=3t^2-3\\ \Rightarrow \frac{dy}{dt}=3\left(t^2-1\right)\end{array}$

  $\frac{dx}{dt}=\frac{d}{dt}\left(t^2\right)$

Use derivative rule $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

  $\Rightarrow \frac{dx}{dt}=2t$

  $\frac{dy}{dx}\left(\text{slope}\right)=\frac{3\left(t^2-1\right)}{2t}$

For the horizontal tangent line.

  $\begin{array}{l}\frac{dy}{dx}=0\\ \Rightarrow \frac{3\left(t^2-1\right)}{2t}=0\\ \Rightarrow 3\left(t^2-1\right)=0\\ \Rightarrow t^2-1=0\\ \Rightarrow t^2=1\\ \Rightarrow t=\pm \sqrt{1}\\ \Rightarrow t=\pm 1\end{array}$

Substitute $t=-1$ in the given equations.

  $\begin{array}{l}x=t^2={\left(-1\right)}^2=1\\ y=t^3-3t={\left(-1\right)}^3-3\cdot \left(-1\right)=2\\ \left(x,y\right)=\left(1,2\right)\end{array}$

Substitute $t=1$ in the given equations.

  $\begin{array}{l}x=t^2={\left(1\right)}^2=1\\ y=t^3-3t={\left(1\right)}^3-3\cdot \left(1\right)=-2\\ \left(x,y\right)=\left(1,-2\right)\end{array}$

For vertical tangent line $\frac{dx}{dt}=0$ .

  $\begin{array}{l}\frac{dx}{dt}=0\\ \Rightarrow 2t=0\\ \Rightarrow t=0\end{array}$

Substitute $t=0$ in the given equations.

  $\begin{array}{l}x=t^2=0^2=0\\ y=t^3-3t={\left(0\right)}^3-3\cdot \left(0\right)=0\\ \left(x,y\right)=\left(0,0\right)\end{array}$

Hence the tangents are horizontal at the points $\left(2,2\right)$ and $\left(2,-2\right)$ . The tangent is vertical at the point $\left(0,0\right)$ .

c.

To determine

Draw a graph for the part $\left(a\right)$ and part $\left(b\right)$ .

Explanation of Solution

Given:

The given equations are $x=t^2$ and $y=t^3-3t$ . The given point is $\left(3,0\right)$ .

Calculation:

Find the slope.

Apply formula $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ .

  $\frac{dy}{dt}=\frac{d}{dt}\left(t^3-3t\right)$

Apply difference rule $\left(f-g\right)\text{'}=f\text{'}-g\text{'}$ .

  $\Rightarrow \frac{dy}{dt}=\frac{d}{dt}\left(t^3\right)-\frac{d}{dt}\left(3t\right)$

Use derivative rule $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

  $\begin{array}{l}\Rightarrow \frac{dy}{dt}=3t^2-3\\ \Rightarrow \frac{dy}{dt}=3\left(t^2-1\right)\end{array}$

  $\frac{dx}{dt}=\frac{d}{dt}\left(t^2\right)$

Use derivative rule $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

  $\Rightarrow \frac{dx}{dt}=2t$

  $\frac{dy}{dx}\left(\text{slope}\right)=\frac{3\left(t^2-1\right)}{2t}$

For the horizontal tangent line.

  $\begin{array}{l}\frac{dy}{dx}=0\\ \Rightarrow \frac{3\left(t^2-1\right)}{2t}=0\\ \Rightarrow 3\left(t^2-1\right)=0\\ \Rightarrow t^2-1=0\\ \Rightarrow t^2=1\\ \Rightarrow t=\pm \sqrt{1}\\ \Rightarrow t=\pm 1\end{array}$

Substitute $t=-1$ in the given equations.

  $\begin{array}{l}x=t^2={\left(-1\right)}^2=1\\ y=t^3-3t={\left(-1\right)}^3-3\cdot \left(-1\right)=2\\ \left(x,y\right)=\left(1,2\right)\end{array}$

Substitute $t=1$ in the given equations.

  $\begin{array}{l}x=t^2={\left(1\right)}^2=1\\ y=t^3-3t={\left(1\right)}^3-3\cdot \left(1\right)=-2\\ \left(x,y\right)=\left(1,-2\right)\end{array}$

For vertical tangent line $\frac{dx}{dt}=0$ .

  $\begin{array}{l}\frac{dx}{dt}=0\\ \Rightarrow 2t=0\\ \Rightarrow t=0\end{array}$

Substitute $t=0$ in the given equations.

  $\begin{array}{l}x=t^2=0^2=0\\ y=t^3-3t={\left(0\right)}^3-3\cdot \left(0\right)=0\\ \left(x,y\right)=\left(0,0\right)\end{array}$

Draw a table for the curve $C$ .

$t$	$x=t^2$	$y=t^3-3t$
$\begin{array}{l}0\\ 1\\ 2\\ 3\\ 4\end{array}$	$\begin{array}{l}0\\ 1\\ 4\\ 9\\ 16\end{array}$	$\begin{array}{l}0\\ -2\\ 2\\ 18\\ 52\end{array}$

Hence the graph of the curve is given below.