Question

Derivative with respect to t of:
V=(4/3)pi (sqrt(a)/sqrt(4pi))^3

Could you please write the steps? I want to make sure I am not messing up my work.

Answer 1

That's not a function of t, so it won't change with respect to t.

Answer 2

It's for a related rates problem.

Answer 3

Alright so you need to give me something with a "t" in it in order to take the derivative with respect to t, otherwise it's a constant and the derivative is just 0.

Answer 4

Maybe tell me the context or something? You have an "a" in there, so I could show you how to take the derivative with respect to a.

Answer 5

I substituted the surface area of a sphere into the equation to the volume of a sphere. I had
\[A=4pir^2 \rightarrow r=\sqrt{\frac{ A }{ 4\pi }}\]
Then substituted that into the r for the volume.
\[V=\frac{ 4\pi }{ 3 }\left( \frac{ \sqrt{A} }{ \sqrt{4\pi} } \right)^3\]

The actual question is:
A skier stands at the edge of a ski-drop and decides to make a snowball, then he rolls the snowball down the slope. He measures that the volume of the snowball increased by 4 cm3 during the 30 seconds it rolled down the slope. At what rate will the surface area of the snowball be increasing when the surface area is 10 m2.

Answer 6

I get the following:
\[\frac{ dV }{ dt }=\frac{ \sqrt{A} }{ 4\sqrt{\pi} }\frac{ dA }{ dt }\]
but I'm not 100% sure about it.

Answer 7

your problem is not easy, however, I think we should find out dr /dt, since both V and A relate to r , so from \(\dfrac{\triangle V} {\triangle t}\) can we find out the rate r respect to t?? and from that function of r, we replace to A ?
@dan815

Answer 8

Are you sure? At this point all I'd have to do is plug in my known values to get my dA/dt but I am unsure if I did my implicit differentiation correctly when relating the two formulas.

Answer 9

hehehehe... I have to confess that I am not sure. That's why I tag dan. He is much more better than me.

Answer 10

Ah, alright no problem. I appreciate any help I can get.

Answer 11

@zepdrix

Answer 12

I sort of like the spirit of this question, however I feel like they might be leaving some things out here. Are we to assume that it is gaining volume at a constant rate? Or as it rolls down the hill, does it speed up and gain volume in proportion to the amount of surface area it has? I suppose I'm over thinking it. So without further ado, I'll show you how to solve this based on this being simplified.

Answer 13

Yeah the rate at which the volume increases will be constant, don't take into account the speed gained from the gained mass and such.

Answer 14

First off, what do we know? We know that over a change of time it changes volume. So we effectively have: dV/dt, since we know it gained 4 cm^3 in 30 seconds.

But we want to know dS/dt when it has a surface area of 10m^2. So we can leapfrog our way there by seeing:
\[\frac{ dV }{ dt } \frac{ dS }{ dV }=\frac{ dS }{ dt }\]
Since we know dV/dt we simply multiply this by the derivative of surface area with respect to volume. Seems weird, but you'll just have to plug one into the other like I believe you did earlier (didn't check) and you'll get dS/dV from this simple little thing: \[S=4 \pi( \frac{ 3V }{ 4 \pi })^{2/3}\]

But you'll get dS/dV in terms of V. You can then just plug in the other formula or find what the volume is when the surface area is 10 m^2 so that dV/dt * dS/dV tells you dS/dt at the surface area! Voila!

Answer 15

If you want me to show a few more steps, you'll have to show me your steps and guesses and I'll help you narrow it down to fix what you need.

Answer 16

The actual question is:
A skier stands at the edge of a ski-drop and decides to make a snowball, then he rolls the snowball down the slope. He measures that the volume of the snowball increased by 4 cm3 during the 30 seconds it rolled down the slope. At what rate will the surface area of the snowball be increasing when the surface area is 10 m2.