Chapter 3.5, Problem 30E

(a)

To determine

To find: The equation of the tangent line to the given equation at the point.

Answer to Problem 30E

The equation of the tangent line to the equation $y^2=x^3+3x^2$ at the point $\left(1,-2\right)$ is $9x+4y=1$.

Explanation of Solution

Given:

The curve with equation $y^2=x^3+3x^2$.

The point is $\left(1,-2\right)$.

Derivative rules:

(1) Chain rule: If $y=f\left(u\right)$ and $u=g\left(x\right)$  are both differentiable function, then

$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$.

(2) Product rule: If $y=f\left(u\right)$ and $u=g\left(x\right)$  are both differentiable function, then

$\frac{d}{dx}\left(f\left(x\right)+g\left(x\right)\right)=\frac{d}{dx}\left(f\left(x\right)\right)+\frac{d}{dx}\left(g\left(x\right)\right)$.

Formula used:

The equation of the tangent line at $\left(x_1,y_1\right)$ is, $y-y_1=m\left(x-x_1\right)$ (1)

Here, m is the slope of the tangent line at $\left(x_1,y_1\right)$ and ${m=\frac{dy}{dx}|}_{x=x_1,y=y_1}$.

Calculation:

Obtain the equation of tangent line to the given point.

$y^2=x^3+3x^2$

Differentiate the above equation implicitly with respect to x,

$\begin{array}{l}\frac{d}{dx}\left(y^2\right)=\frac{d}{dx}\left(x^3+3x^2\right)\\ \frac{d}{dx}\left(y^2\right)=\frac{d}{dx}\left(x^3\right)+3\frac{d}{dx}\left(x^2\right)\\ \frac{d}{dx}\left(y^2\right)=3x^2+6x\end{array}$

Apply the chain rule (1) and simplify the terms,

$\begin{array}{l}\frac{d}{dy}\left(y^2\right)\frac{dy}{dx}=3x^2+6x\\ 2y\frac{dy}{dx}=3x^2+6x\\ \frac{dy}{dx}=\frac{3x^2+6x}{2y}\end{array}$

Therefore, the derivative of the equation is $\frac{dy}{dx}=\frac{3x^2+6x}{2y}$.

The slope of the tangent line at $\left(1,-2\right)$ is computed as follows,

$\begin{array}{c}m={\frac{dy}{dx}|}_{\left(x_1,y_1\right)=\left(1,-2\right)}\\ =\frac{3{\left(1\right)}^2+6\left(1\right)}{2\left(-2\right)}\\ =-\frac{9}{4}\end{array}$

Thus, the slope of the tangent line at $\left(1,-2\right)$ is $m=-\frac{9}{4}$.

Substitute $\left(1,-2\right)$ for $\left(x_1,y_1\right)$ and $m=-\frac{9}{4}$ in equation (1),

$\begin{array}{c}y+2=\frac{-9}{4}\left(x-1\right)\\ y+2=\frac{-9x}{4}+\frac{9}{4}\\ \frac{4y+8}{4}=\frac{-9x}{4}+\frac{9}{4}\\ 9x+4y=1\end{array}$

Therefore, the equation of the tangent line to the equation $y^2=x^3+3x^2$ at the point $\left(1,-2\right)$ is $9x+4y=1$.

(b)

To determine

To find: The points if the curve has horizontal tangent.

Answer to Problem 30E

The curve has horizontal tangent at $\left(-2,-2\right)\text{ and }\left(-2,2\right)$.

Explanation of Solution

Given:

The curve with equation $y^2=x^3+3x^2$.

The derivative of the equation $\frac{dy}{dx}=\frac{3x^2+6x}{2y}$.

Calculation:

Obtain the point if the curve has horizontal tangent.

Note that, the curve horizontal tangent if $\frac{dy}{dx}=0$.

Suppose that $\frac{dy}{dx}=0$,

$\begin{array}{l}\frac{3x^2+6x}{2y}=0\\ 3x^2+6x=0\\ x\left(3x+6\right)=0\\ x=0\text{ or }x=-2\end{array}$

Rewrite the given equation as follows,

$y=\sqrt{x^3+3x^2}$

Substitute $x=0$ in the above equation,

$\begin{array}{c}y=\sqrt{0^3+3{\left(0\right)}^2}\\ =0\end{array}$

The derivative of the equation does not exist at $\left(0,0\right)$.

Substitute $x=-2$ in $y=\sqrt{x^3+3x^2}$,

$\begin{array}{c}y=\sqrt{{\left(-2\right)}^3+3{\left(-2\right)}^2}\\ =\sqrt{-8+12}\\ =\sqrt{4}\\ =\pm 2\end{array}$

Therefore, the curve has horizontal tangent at $\left(-2,-2\right)\text{ and }\left(-2,2\right)$.

(c)

To determine

To sketch: The curve and tangent line to the given point.

Explanation of Solution

Graph:

Using online graphic calculator to draw the curve and the tangent line as shown below in Figure 1,

From Figure 1, it is observed that the line $9x+4y=1$ touch the curve $y^2=x^3+3x^2$ at $\left(1,-2\right)$. That is, the tangent line $9x+4y=1$ to the curve at $\left(1,-2\right)$ and the curve has horizontal tangent at $\left(-2,-2\right)\text{ and }\left(-2,2\right)$.