Chapter 4.2, Problem 47E

(a)

To determine

To calculate: To confirm that the point $\left(-1,1\right)$ lies on the curve.

Answer to Problem 47E

The point lies on the curve.

Explanation of Solution

Given Information: The equation of the curve is $x^3{y}^2=\cos \left(\pi y\right)$

Calculation:

Substitute the point $\left(-1,1\right)$ in the curve,

  $\begin{array}{c}{x}^3{y}^2=\cos \left(\pi y\right)\\ {\left(-1\right)}^3{\left(1\right)}^2=\cos \left(\pi \times 1\right)\left(\text{Substitute }x=-1,y=1\right)\\ -1\times 1=\cos \pi \\ -1=-1\left(\because \cos \pi =-1\right)\end{array}$

Thus, the values on both the sides are same. Therefore, the point lies on the curve.

The point lies on the curve.

(b)

To determine

To calculate: To find the slope of the line tangent to the curve at the point $\left(-1,1\right)$

Answer to Problem 47E

The slope of the line tangent to the curve at the point $\left(-1,1\right)$ is $\frac{3}{2}$ .

Explanation of Solution

Given Information: The equation of the curve is $x^3{y}^2=\cos \left(\pi y\right)$

Calculation:

The slope of the line tangent to the curve is,

  $\begin{array}{c}{x}^3{y}^2=\cos \left(\pi y\right)\\ \frac{d}{dx}\left(x^3{y}^2\right)=\frac{d}{dx}\left(\cos \left(\pi y\right)\right)\\ x^3\frac{d}{dx}\left(y^2\right)+y^2\frac{d}{dx}\left(x^3\right)=\frac{d}{dx}\left(\cos \left(\pi y\right)\right)\\ x^3\left(2y\frac{dy}{dx}\right)+y^2\left(3x^2\right)=0\\ 2x^{3}y\frac{dy}{dx}+3x^2{y}^2=0\\ 2x^{3}y\frac{dy}{dx}=-3x^2{y}^2\\ \frac{dy}{dx}=-\frac{3x^2{y}^2}{2x^{3}y}\\ =-\frac{3y}{2x}\end{array}$

Substitute the point $\left(-1,1\right)$ in the curve,

  $\begin{array}{c}{\left[\frac{dy}{dx}\right]}_{\left(-1,1\right)}={\left[-\frac{3y}{2x}\right]}_{\left(-1,1\right)}\\ =-\frac{3\times 1}{2\times -1}\\ =\frac{3}{2}\end{array}$

Hence, the slope of the line tangent to the curve at the point $\left(-1,1\right)$ is $\frac{3}{2}$ .

The slope of the line tangent to the curve at the point $\left(-1,1\right)$ is $\frac{3}{2}$ .