Question

Find the value of `(dr)/dt` at `t=0` if `r=(theta^2 +7)^(1/3)` and `theta^2t + theta =1`?

So far, I have

`(dr)/(d theta) = 1/3(theta^2+7)^(-2/3) * 2theta`

and

`(d theta)/dt = (2thetat + 1) / -theta^2`

I tried to multiply them together, but that doesn't get me anywhere.

Answer 1

`-1/6`

Explanation:

You're definitely on the right track!

As you pointed out, we start with chain rule:
`(dr)/(dt) = (dr)/(d theta) * (d theta)/(dt)`

You calculated these values correctly.

At `t=0`, the second equation gives `theta = 1`. The second equation then gives that `r = 2` at that time. So we can actually plug those values into the equations you supplied, yielding
`(dr)/(d theta) = 1/3 * (8)^(-2/3) * 2 * 1 = 2/3 * 1/4 = 1/6`

`(d theta)/dt = (2 * 1 * 0 + 1)/(-1^2) = -1`

Hence,
`(dr)/(dt) = (dr)/(d theta) * (d theta)/(dt) = -1/6`.