Another related rate

sigma

Junior Member
Joined
Feb 19, 2006
Messages
106
Just had this one on an exam.

The surface area of a cube is increasing at a rate 4 meters squared/sec. How much is the volume of the cube increasing when the length of the cube is 10?


So a picture of a cube with a side labeld as 10.
Then figuring out what rates I have and what I needed, there's what I did. Am I right? I don't think so.

\(\displaystyle \
\L\begin{array}{l}
\frac{{dsa}}{{dt}} = 4m^2 /\sec \\
\frac{{dv}}{{dt}} = ??? \\
\frac{{da}}{{dt}} = ??? \\
\to sa = 6a \\
\to \frac{{dsa}}{{dt}} = 6\frac{{da}}{{dt}} \\
\to 4 = 6\frac{{da}}{{dt}} \\
\to \frac{{da}}{{dt}} = \frac{2}{3} \\
\to v = a^3 \\
\to \frac{{dv}}{{dt}} = 3a^2 \frac{{da}}{{dt}} \\
\to {\rm when a = }10x10x10 = 1000, \\
\to \frac{{dv}}{{dt}} = 3(1000)^2 \frac{2}{3} \\
\to \frac{{dv}}{{dt}} = 1000^2 (2) \\
\to \frac{{dv}}{{dt}} = 20,000,000!!! \\
\end{array}
\
\\)

Dear god I messed up on this question! How do you figure out related rates involving cubes? One of the only types of questions we did not go over before the exam. The only real probelm is I couldn't remember what the surface area or volume of a cube was (the formulas). I semi guessed the surface area was equal to 6 times the total area because a cube has 6 sides and that the volume of a cube is equale to the area cubed but then again, I don't think thats right. Then trying to figure out how to relate surface area to volume I could not remember.
 
Let's give this a try sigma.

Sorry for the flub up. After I posted I thought something was amiss, but it

was too late to log back on.
 
In other words, this has nothing to do with the area at all. I almost had the equations set up right for surface area and volume. Oh well, half to quarter marks I guess.

Just for fun (or a redundant question) galactus, if you were marking this question and the question was worth 8 marks, how many marks would you give me for my work?:?
 
galactus, I'm curious how you've done this... Maybe if you explained your steps a bit more clearly?

Especially the 3(S/6)dS/dt. Not sure where that came from...

Thanks.
 
I'm not sure, LaTex looks so convincing! I would do it as

S=6x²
dS/dt = 12x dx/dt = 4
dx/dt = 1/(3x)

V=x^3
dV/dx =3x^2

dV/dt = dV/dx*dx/dt =
3x^2*(1/3x) =
x
 
I'm back to fix my previous mistake. I made a bad substitution.

Find \(\displaystyle \L\\\frac{dV}{dt}\) given \(\displaystyle \L\\\frac{dS}{dt}=4\)

\(\displaystyle \L\\S=6x^{2}\)

\(\displaystyle \L\\V=x^{3}\)

\(\displaystyle \L\\x=\sqrt{\frac{S}{6}}\)

\(\displaystyle \L\\V=(\sqrt{\frac{S}{6}})^{3}=\frac{(6S)^{\frac{3}{2}}}{216}\)

\(\displaystyle \L\\\frac{dV}{dt}=\frac{6^{\frac{3}{2}}\sqrt{S}}{144}\frac{dS}

{dt}=\frac{\sqrt{6S}}{24}=\frac{\sqrt{6(6x^{2})}}{24}=\frac{x}{4}\frac{dS}{dt}\)

\(\displaystyle \L\\\frac{x}{4}\frac{dS}{dt}=\frac{10}{4}(4)=10 m^{3}/sec\)

This is what I was getting at originally and messed up. I believe it agrees with your more efficient method, Gene.
 
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