# Thread: help with derivatives (applied problem)

1. ## help with derivatives (applied problem)

Can anyone help me figure this problem out?

Any help is appreciated! Thanks

2. What have you tried so far?

3. i answered part a with V=s^3 A=s^2...i dont know what it means to differentiate dV/dt with respect to s...

4. Originally Posted by colerelm1
i answered part a with V=s^3 A=s^2...i dont know what it means to differentiate dV/dt with respect to s...
Good!

$\displaystyle \frac{dV}{dt}=kA=ks^2$

Do you think k will be positive or negative? (Remember that the volume is decreasing with time.)

5. Did you mean "differentiate $V$ with respect to $t$? If so, then it's called implicit differentiation:

$\dfrac{dV}{dt} = 3s^2 \dfrac{ds}{dt}$

6. Originally Posted by alexmahone
Good!

$\displaystyle \frac{dV}{dt}=kA=ks^2$

Do you think k will be positive or negative? (Remember that the volume is decreasing with time.)
ok i know its going to be negative because its decreasing but how do I do part c? I honestly have no clue where to start this one. My thinking is that dV/dt means volume(y axis) with respect to time (x axis) related to some ds/dt where s (y axis) is being differentiated with respect to time (x axis). I know how to read leibniz notation I just can't seem to grasp an understanding of it...

7. Originally Posted by colerelm1
ok i know its going to be negative because its decreasing but how do I do part c? I honestly have no clue where to start this one. My thinking is that dV/dt means volume(y axis) with respect to time (x axis) related to some ds/dt where s (y axis) is being differentiated with respect to time (x axis). I know how to read leibniz notation I just can't seem to grasp an understanding of it...
$\displaystyle V=s^3$

Using the Chain Rule,

$\displaystyle \frac{dV}{dt}=\frac{dV}{ds}\frac{ds}{dt}=3s^2\frac {ds}{dt}$

8. Originally Posted by alexmahone
$\displaystyle V=s^3$

Using the Chain Rule,

$\displaystyle \frac{dV}{dt}=\frac{dV}{ds}\frac{ds}{dt}=3s^2\frac {ds}{dt}$
ok so for part (d) i got the answer of:

$ks^2 = 3s^2 ds/dt$
which is the same as $ds/dt = ks^2 / 3s^2$

Now, for part (e) I don't get how I would use my answer from (d) to write an equation which relates s(0) to s(1) and then use it to find t_melt in terms of the quantity s_1/s_0

Can anyone help me out?