# Thread: Rates of change - surface area vs. volume

1. ## Rates of change - surface area vs. volume

At first this question seemed easy, however I still can't get the answer. I've been using the formula:

dx/dt = dx/dy * dy/dt

The surface area of a cube is changing at the rate of 8cm^2/second. How fast is the volume changing when the surface area is 60 cm?

2. Originally Posted by freswood
At first this question seemed easy, however I still can't get the answer. I've been using the formula:

dx/dt = dx/dy * dy/dt

The surface area of a cube is changing at the rate of 8cm^2/second. How fast is the volume changing when the surface area is 60 cm?

$\displaystyle S=6a^2=60cm^2$
$\displaystyle \frac{dS}{dt}=12a\frac{da}{dt}=8cm^2/s$
$\displaystyle \frac{dV}{dt}=3a^2\frac{da}{dt}$
Get the value of a from first equation
Divide equation 3 by equation 2

KeepSmiling
Malay

3. I'm quite confused

What is dx/dy? I'm not really sure how to relate surface area and volume. And what's the final answer? I want to see if you get the same answer as what's in the book.

4. Originally Posted by freswood
I'm quite confused

What is dx/dy? I'm not really sure how to relate surface area and volume. And what's the final answer? I want to see if you get the same answer as what's in the book.
I think the answer is $\displaystyle 2\sqrt{10}$

KeepSmiling
Malay

5. I'm still not sure where you got those values from.

dx/dt = 8, but what is dx/dy and how did you get it? Is it a formula for surface area with respect to volume?

6. Originally Posted by freswood
I'm still not sure where you got those values from.

dx/dt = 8, but what is dx/dy and how did you get it? Is it a formula for surface area with respect to volume?
I am sorry but I don't want to give you a solution straightaway.
Answer these simple questions and solve the problem yourself.
1)What do you intend to find?Write down an expression for it.
2)I am listing a few things. Tell whether you can find their value, if yes, what is their value:
S
a
V
dS/dt
da/dt
dS/da
dV/da
dS/dV
dV/dS
3)Write an expression for dV/dt. What values do you need for it?Can you find those values in ques.2?
I hope that you can solve the problem now. I still you have any problem, post it and i will try to tell you, straightaway this time.
To find dy/dx . There are two cases(for explicit functions):you want to find dy/dx in terms or yuo want to find dy/dx in terms of y.
To find in terms of x, express y as a function of x and differentiate.
to find in terms of y, express x as a function of y and differentiate.

Keep Smiling
Malay

7. Originally Posted by malaygoel
I am sorry but I don't want to give you a solution straightaway.
Answer these simple questions and solve the problem yourself.
1)What do you intend to find?Write down an expression for it.
2)I am listing a few things. Tell whether you can find their value, if yes, what is their value:
S
a
V
dS/dt
da/dt
dS/da
dV/da
dS/dV
dV/dS
3)Write an expression for dV/dt. What values do you need for it?Can you find those values in ques.2?
I hope that you can solve the problem now. I still you have any problem, post it and i will try to tell you, straightaway this time.
To find dy/dx . There are two cases(for explicit functions):you want to find dy/dx in terms or yuo want to find dy/dx in terms of y.
To find in terms of x, express y as a function of x and differentiate.
to find in terms of y, express x as a function of y and differentiate.

Keep Smiling
Malay
1) dV/dt
2)
S = 6a^2 where a is side length
a = 10^.5
V = a^3
dS/dt = 8
da/dt
dS/da = 12a
dV/da = 3a^2
dS/dV
dV/dS

3) For dV/dt you need dx/dt and dS/dV

But that's the whole problem - I don't know how to find dS/dV. I know you think you're doing the right thing, and I appreciate your help, but I like to be able to work backwards. This question is causing me a lot of grief. I'm doing maths a year ahead, and I find it hard to keep up with the older girls.

8. Originally Posted by freswood
1) dV/dt
2)
S = 6a^2 where a is side length
a = 10^.5
V = a^3
dS/dt = 8
da/dt
dS/da = 12a
dV/da = 3a^2
dS/dV
dV/dS

3) For dV/dt you need dx/dt and dS/dV
No you don't
Now see.
dV/dt = $\displaystyle \frac{dV}{da}*\frac{da}{dt}$
dV/da=3a^2(you know the value of a)
You couldn't find da/dt
You know dS/dt = 8
$\displaystyle \frac{dS}{da}*\frac{da}{dt}$=8
Find the value of da/dt from here and the question is solved.

KeepSmiling
Malay

9. Originally Posted by freswood
At first this question seemed easy, however I still can't get the answer. I've been using the formula:

dx/dt = dx/dy * dy/dt

The surface area of a cube is changing at the rate of 8cm^2/second. How fast is the volume changing when the surface area is 60 cm?

I think (but am probably wrong) the problem here is that x,t,and y
in dx/dt = dx/dy * dy/dt, are serving as place holders for whatever is
relevant to the problem at hand.

For this problem you replace x by V (representing the volume of the
cube) and y by S (representing the surface area of the cube) and let
t represent time. Then your ralation between derivatives becomes:

dV/dt=dV/dS * dS/dt.

You are told dS/dt=8 (cm^2/s), so you need to find dV/dS.

The volume V of a cube of side a is a^3, and the surface area of
such a cube is 6a^2, so expressing V in terms of S gives:

V=(S/6)^(3/2),

and the rest should be plain sailing.

RonL

10. That's brilliant! Thanks so much! It was just that I couldn't understand how to relate surface area to volume, so couldn't come up with the equation dS/dV.