Question

If the rate of change of the function f(t) = t^(4/3) when t=c is twice the rate of change of f(t) when t=8, then c=?

Answer 1

Does the wording even make sense to you?

Answer 2

Yes. First find the derivative of f of t. Do you know what that is?

Answer 3

f'(t)*t' right?

Answer 4

so, (4/3)t^(1/4)

Answer 5

exponet is 1/3

Answer 6

right, typo on my part, thank you.

Answer 7

Now set up an equation involving c and 8\[\frac{ 4 }{ 3 }(c)^{\frac{ 1 }{ 3 }}=(2)\frac{ 4 }{ 3 }(8)^{\frac{ 1 }{ 3 }}\]

Answer 8

how did you get that equation?

Answer 9

Since I knew the formula for each rate of change, I substituted c for t on one side and 8 on the other, just as the problem stipulated. And then I added that the rate of change at c is twice that at 8

Answer 10

Oh, alright, cool

Answer 11

Can you solve for c then?

Answer 12

I got 3root4... am I doing something wrong?

Answer 13

\[\Large f'(t)=\frac{4}{3}t^{\frac{1}{3}}\]
when t=c
\[\Large f'(t)=\frac{4}{3}c^{\frac{1}{3}}\]
now again
\[\Large f'(t)=\frac{4}{3}t^{\frac{1}{3}}\]
when t=8 then
\[\Large f'(t)=\frac{4}{3}8^{\frac{1}{3}}\]
since when its t=c its twice so set the two equal
\[\Large 2(\frac{4}{3})C^{\frac{1}{3}}=\frac{4}{3}*(8)^{\frac{1}{3}}\]
just solve for C

Answer 14

Sort of, but you're close. Reduce the equation to its bare bones. On one side, the right, you have 2 times 4/3 times the cubic root of 8 (or 2). On the other side, you also have 4/3, which can cancel with the 4/3 on the other side

Answer 15

I'm trying to hurry here. \[c ^{\frac{ 1 }{ 3 }}=2*2\]

Answer 16

oh, so 4^3, which is 64

Answer 17

Right

Answer 18

Last equation in my response is
\[\Large \frac{4}{3}C^{\frac{1}{3}}=2 (\frac{4}{3})8^{\frac{1}{3}}\]

Answer 19

Righty, thank you so much! :D