Chapter 11.3, Problem 8QR

To determine

To calculate: the point on the curve where the slop is zero.

Answer to Problem 8QR

The pints on the curve where the slop is zero are $\left(0,5\right)\text{and}\left(0,-5\right)$ .

Explanation of Solution

Given information:

  $x=3\cos t$

  $y=5\sin t$

  $0\le t\le 2\pi$

Formula used:

  $du/dv=\frac{du/dx}{dv/dx}$

Calculation:

Given that theexpression of $x$ is $3\cos t$ for $0\le t\le 2\pi$ .

Take differentiation of $x$ with respect to $t$ .

  $\begin{array}{c}dx/dt=\frac{d}{dt}3\cos t\\ =-3\sin t\end{array}$

Similarly, the given expression of $y$ is $5\sin t$ for $0\le t\le 2\pi$ .

Take differentiation of $y$ with respect to $t$ .

  $\begin{array}{c}dy/dt=\frac{d}{dt}5\sin t\\ =5\cos t\end{array}$

Now, calculate $dy/dx$ . Use the formula $du/dv=\frac{du/dx}{dv/dx}$ .

  $\begin{array}{c}dy/dx=\frac{dy/dt}{dx/dt}\\ =\frac{5\cos t}{-3\sin t}\\ =-\frac{5}{3}\cot t\end{array}$

Given that the slop of the curve is zero.

  $\begin{array}{c}dy/dx=0\\ -\frac{5}{3}\cot t=0\\ \cot t=0\end{array}$

  $\therefore t=\frac{\pi }{2},\frac{3\pi }{2}$

Substitute the value of $t$ in the expression of $x$ and $y$ .

  $\begin{array}{c}x\left(\frac{\pi }{2}\right)=3\cos \left(\frac{\pi }{2}\right)\\ =0\\ x\left(\frac{3\pi }{2}\right)=3\cos \left(\frac{3\pi }{2}\right)\\ =0\end{array}$

Similarly,

  $\begin{array}{c}y\left(\frac{\pi }{2}\right)=5\sin \left(\frac{\pi }{2}\right)\\ =5\\ y\left(\frac{3\pi }{2}\right)=5\sin \left(\frac{3\pi }{2}\right)\\ =-5\end{array}$

Conclusion:

Hence, the required pints on the curve where the slop is zero are $\left(0,5\right)\text{and}\left(0,-5\right)$ .