Question

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

Answer 1

`\left (-(4)/(√(5)),(8)/(√(5))\right )\text{ and }\left ((4)/(√(5)),-(8)/(√(5))\right )`.

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)`

`(dx)/(dt)=-4sin (t)`

`y=4sin (t)`

`(dy)/(dt)=4cos (t)`

Now,

`(dy)/(dx)=(dy)/(dt)* (dt)/(dx)`

`(dy)/(dx)=4cos (t)* (1)/(-4sin (t))`

`(dy)/(dx)=-\cot t`

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

`-\cot t=(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 √(5)`

`\cos t=\pm (1)/(√(5))`

Now,

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

`\sin t=\pm \sqrt{1-(1)/(5)}`

`\sin t=\pm \sqrt{(4)/(5)}`

`\sin t=\pm (2)/(√(5))`

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

If
`\sin t= (2)/(√(5))`, then
`\cos t=- (1)/(√(5))`.

`x=4cos (t)=-(4)/(√(5))`

`y=4sin (t)=(8)/(√(5))`

If
`\sin t=- (2)/(√(5))`, then
`\cos t= (1)/(√(5))`.

`x=4cos (t)=(4)/(√(5))`

`y=4sin (t)=-(8)/(√(5))`

Therefore, the two points are
`\left (-(4)/(√(5)),(8)/(√(5))\right )\text{ and }\left ((4)/(√(5)),-(8)/(√(5))\right )`.