# c4 Differential Equations help please!

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#1
I cant work out this question and the mark scheme doesn't seem to help.. At time t seconds the length of the side of a cube is xcm, the surface area of the cube is Scm^2, and the volume is Vcm^3.
The surface area of the cube is increasing at a constant rate of 8cm^2s^-1.

a. Show that dx/dt = k/x

b. dV/dx=2V^1/3

Help will be greatly appreciated! thanks 0
12 years ago
#2
(a) The side of the cube is of length x cm.
The volume of the cube is given by width x height x depth, V = x^3.

A cube has 6 faces. The area of each face is given by height x width (x^2).
Total surface area = S = 6x^2.

Differentiating the surface area with respect to the length of each side gives dS/dx = 12x.

The question states that the surface area of the cube is increasing by 8cm per second, so the rate of change of the surface area with respect to time is dS/dt = 8.

Now, using the chain rule:

dx/dt = dx/dS x dS/dt
dx/dt = [1/12x]
dx/dt = 2/3x

[k = 2/3]

(b) S = 6x^2

Rearranging to get x in terms of S gives x = (S/6)^1/2

V = x^3
Substituting x in terms of S into the volume gives
V = (S/6)^3/2

Raising both sides to the power of 2/3 gives
V^2/3 = S/6
6V^2/3 = S

6(2/3)V^-1/3 = dS/dV

4V^-1/3 = dS/dV

dV/dt = dV/dS x dS/dt
dV/dt = [V^(1/3)/4]
dV/dt = 2V^1/3

I really hope that's helpful. Was a very tricky question for me, got stuck on it today when doing this past paper.
1
#3
Thanks, that really helped. It seemed so obvious once it was explained. I guess it always does thanks again!
0
11 years ago
#4
i was stuck on this too. thanks v.much.
0
11 years ago
#5
(Original post by ablendofdrei)
(a) The side of the cube is of length x cm.
The volume of the cube is given by width x height x depth, V = x^3.

A cube has 6 faces. The area of each face is given by height x width (x^2).
Total surface area = S = 6x^2.

Differentiating the surface area with respect to the length of each side gives dS/dx = 12x.

The question states that the surface area of the cube is increasing by 8cm per second, so the rate of change of the surface area with respect to time is dS/dt = 8.

Now, using the chain rule:

dx/dt = dx/dS x dS/dt
dx/dt = [1/12x]
dx/dt = 2/3x

[k = 2/3]

(b) S = 6x^2

Rearranging to get x in terms of S gives x = (S/6)^1/2

V = x^3
Substituting x in terms of S into the volume gives
V = (S/6)^3/2

Raising both sides to the power of 2/3 gives
V^2/3 = S/6
6V^2/3 = S

6(2/3)V^-1/3 = dS/dV

4V^-1/3 = dS/dV

dV/dt = dV/dS x dS/dt
dV/dt = [V^(1/3)/4]
dV/dt = 2V^1/3

I really hope that's helpful. Was a very tricky question for me, got stuck on it today when doing this past paper.
Why can't you just do dV/dx = 1/dx/dV and then differentiate x=V^(1/3)? I ask this because it gives the wrong answer, of 3V^(2/3) :/
0
11 years ago
#6
what past paper is tht of?
0
11 years ago
#7
(Original post by Toneh)
Why can't you just do dV/dx = 1/dx/dV and then differentiate x=V^(1/3)? I ask this because it gives the wrong answer, of 3V^(1/3) :/
i would like to know too 0
1 month ago
#8
Find the general solution for V in terms of t. add on in this question....can u help me solve this
0
1 month ago
#9
(Original post by ablendofdrei)
(a) The side of the cube is of length x cm.
The volume of the cube is given by width x height x depth, V = x^3.

A cube has 6 faces. The area of each face is given by height x width (x^2).
Total surface area = S = 6x^2.

Differentiating the surface area with respect to the length of each side gives dS/dx = 12x.

The question states that the surface area of the cube is increasing by 8cm per second, so the rate of change of the surface area with respect to time is dS/dt = 8.

Now, using the chain rule:

dx/dt = dx/dS x dS/dt
dx/dt = [1/12x]
dx/dt = 2/3x

[k = 2/3]

(b) S = 6x^2

Rearranging to get x in terms of S gives x = (S/6)^1/2

V = x^3
Substituting x in terms of S into the volume gives
V = (S/6)^3/2

Raising both sides to the power of 2/3 gives
V^2/3 = S/6
6V^2/3 = S

6(2/3)V^-1/3 = dS/dV

4V^-1/3 = dS/dV

dV/dt = dV/dS x dS/dt
dV/dt = [V^(1/3)/4]
dV/dt = 2V^1/3

I really hope that's helpful. Was a very tricky question for me, got stuck on it today when doing this past paper.
Find the general solution for V in terms of t.
0
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