Question

Find all points on the curve x=4cos(t),y=4sin(t) that have the slope of 12.

Answer 1

Answer:

$\left (-\dfrac{4}{\sqrt{5}},\dfrac{8}{\sqrt{5}}\right )\text{ and }\left (\dfrac{4}{\sqrt{5}},-\dfrac{8}{\sqrt{5}}\right )$.

Step-by-step explanation:

We need to find all the points on the curve x=4cos(t),y=4sin(t) that have the slope of 1/2.

$x=4cos (t)$

$\dfrac{dx}{dt}=-4sin (t)$

$y=4sin (t)$

$\dfrac{dy}{dt}=4cos (t)$

Now,

$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}$

$\dfrac{dy}{dx}=4cos (t)\times \dfrac{1}{-4sin (t)}$

$\dfrac{dy}{dx}=-\cot t$

So, slope of the curve is $-\cot t$.

$-\cot t=\dfrac{1}{2}$

$-\tan t=2$

$\tan t=-2$            ...(1)

Using $\sec^2t=1+\tan^2t$, we get

$\sec^2t=1+(-2)^2$

$\sec^2t=1+4$

$\sec t=\pm \sqrt{5}$

$\cos t=\pm \dfrac{1}{\sqrt{5}}$

Now,

$\sin^2t=1-cos^2t$

$\sin t=\pm \sqrt{1-\dfrac{1}{5}}$

$\sin t=\pm \sqrt{\dfrac{4}{5}}$

$\sin t=\pm \dfrac{2}{\sqrt{5}}$

It equation (1), tan(t) is negative. So, sin and cos have different signs.

If $\sin t= \dfrac{2}{\sqrt{5}}$, then $\cos t=- \dfrac{1}{\sqrt{5}}$.

$x=4cos (t)=-\dfrac{4}{\sqrt{5}}$

$y=4sin (t)=\dfrac{8}{\sqrt{5}}$

If $\sin t=- \dfrac{2}{\sqrt{5}}$, then $\cos t= \dfrac{1}{\sqrt{5}}$.

$x=4cos (t)=\dfrac{4}{\sqrt{5}}$

$y=4sin (t)=-\dfrac{8}{\sqrt{5}}$

Therefore, the two points are $\left (-\dfrac{4}{\sqrt{5}},\dfrac{8}{\sqrt{5}}\right )\text{ and }\left (\dfrac{4}{\sqrt{5}},-\dfrac{8}{\sqrt{5}}\right )$.