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help with rates of change - c4

Hi i need help with these questions.

these are in exercise 4E of the c4 heinemann book. Q2, Q9, Q10, Q11

Q10 says:

At time t seconds the surface area of a cube is A cm^2 and the volume is V cm^3.

The surface area of the cube is expanding at a constant rate 2 cm^2 s^-1.

Show that dV/dt = 1/2 V^(1/3)




please show me how to do this or give hints. Thanks your help is appreciated and i will rep.
Reply 1
Well here are some equations you might find useful:

V = x^3 ...(i)

A = 6x^2 ...(ii)

dV/dt = dV/dA * dA/dt

Tip: you need to relate A and V using (i) and (ii)
Reply 2
dV/dt = dV/dA dA/dt = dV/dx dx/dA dA/dt

but (dx/dA) = (dA/dx)^-1

so you just differentiate A=6xx and V=xxx, invert the former, and multiply it all together.
Can someone please do it step by step..because I still cannot get it :frown: i am a slow learner..
(edited 6 years ago)
Oh this is an ancient post originally ... in that case

V=x3,A=6x2 V = x^3, A = 6x^2

dVdt=dVdxdxdt {dV \over dt} = {dV \over dx}{dx \over dt}

dAdt=dAdxdxdt=2 {dA \over dt} = {dA \over dx}{dx \over dt} = 2

Can you see how to finish it from here by supplying some values for the derivatives and substituting for dxdt{dx \over dt}?
(edited 6 years ago)
I still can't get the solution. It would be more helpful if full workings and the answer are provided. Thank you so much.
Original post by Adeline Chin
I still can't get the solution. It would be more helpful if full workings and the answer are provided. Thank you so much.


What part don't you get?
I dont get the part to show that dV/dt= 1/2(V)^1/3
The forum ethos is not to give solutions but hints to allow people to solve the problems themselves. You don't seem to be making much effort in that direction. I've already told you pretty explicitly how to do it.

Look at the two equations I've given for derivatives of the Area and Volume with respect to time. Replace any derivative expressions you can. What do the two equations look like now?
I've already gt it. Thank you.
OK good. Well in that case I'll give a more direct solution that avoids the introduction of a variable representing the side length of the cube.

V=(A6)32V = ({A \over 6})^{3 \over 2}

dVdt=32(A6)1216dAdt=12(A6)12=12V13 {dV \over dt} = {3 \over 2} ({A \over 6})^{1 \over 2} {1 \over 6} {dA \over dt} = {1 \over 2} ({A \over 6})^{1 \over 2} = {1 \over 2} V^{1 \over 3}

Probably not what the examiners want to see though.
(edited 6 years ago)