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My Working

Following the hint + some help I got from tutorial below is what I got ... but I believe I did something wrong ... its not the answer below the question yet $0.955\text{ }m\text{ }min$ (I believe its $0.955 m/min$)

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In general how do I start with such problems without any hints? I suppose I must somehow formulate the equation like:

$$\frac{dX}{dt} = \frac{dX}{dY} \cdot \frac{dY}{dt}$$

So that the $dY$ cancels. But suppose the following question ...


I did:

$$\frac{dV}{dt} = 5$$

$$Find \frac{dA}{dt} \text{ when } V = 216 cm^3$$

$$\frac{dV}{dt} = \frac{dA}{dt} \cdot \frac{dV}{dA}$$

So I tried expressing $V$ in terms of $A$: But I still ended up with an $l$

$$V = l^3, A = 6l^2$$

$$V = \frac{1}{6} A l$$


There are a couple of mistakes, at least one of which is a minor slip. I will use the notation of the post.

The equation $$V=\frac{1}{3}\pi h^3 \tan^2\theta$$ for $V$ as a function of $h$ is correct. Since $\tan\theta=\frac{1}{3}$, it would be cleaner to write $V=\frac{1}{27}\pi h^3$. But we will continue using $\tan^2\theta$. Now we differentiate both sides with respect to $t$. Here there are two errors in the post. By the Chain Rule, we have $$\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}=(\pi h^2\tan^2\theta)\frac{dh}{dt}.$$ There was a little slip here, you wrote $2h^2$ for the derivative of $h^3$ with respect to $h$, and it should be $3h^2$. In addition, the necessary $\frac{dh}{dt}$ part, though mentioned correctly once, later appears on the "wrong" side of the equation for $\frac{dV}{dt}$. The Chain Rule is usually an essential tool in related rates problems, and needs to be handled carefully.

Now that we have a general relationship between $\frac{dV}{dt}$ and $\frac{dh}{dt}$, "freeze" things at the instant $t$ when $r=2$. When $r=2$, we have $h=6$. And while the cone is filling up, $\frac{dV}{dt}=12$. So at the instant when $r=2$, we have $$12=(\pi)(6^2)(\tan^2\theta)\frac{dh}{dt}.$$

You did some unnecessary work at the corresponding point of your calculation, but it did not result in an error. You found $\theta$ using the calculator, and then $\tan^2\theta$. That is not needed, since you already know that $\tan\theta=\frac{1}{3}$. Whatever approach we take, we should get $$12=(\pi)(6^2)(1/3)^2\frac{dh}{dt} =4\pi\frac{dh}{dt}.$$ Solve for $\frac{dh}{dt}$. We find that, at the instant when $r=2$, $$\frac{dh}{dt}=\frac{12}{4\pi}=\frac{3}{\pi}.$$ Numerically, the answer is about $0.95493$.

Added: We look briefly at the "cube" problem that was added to the post. We have $\frac{dV}{dt}=5$, and want $\frac{dA}{dt}$. I think the natural thing to do is to let $s=s(t)$ be the side at time $t$. (In the post it is called $l$, fine too, but looks too much like $1$!) Everybody knows that $$V=s^3 \text{ and } A=6s^2$.$$ Don't think, differentiate with respect to time. $$\frac{dV}{dt}=\frac{dV}{ds}\frac{ds}{dt}=3s^2\frac{ds}{dt} \text{ and } \frac{dA}{dt}=12s\frac{ds}{dt}.$$ We have $\frac{dA}{dt}=5$. When $V=216$, $s=6$. From the expression for $\frac{dV}{dt}$ above, we find that $\frac{ds}{dt}=5/108$ when $s=6$. Now we can use the expression for $\frac{dA}{dt}$ to conclude that $\frac{dA}{dt}=(12)(6)(5/108)$. Simplify if desired.

Or else find a direct relationship between $V$ and $h$. We have $V=r^3$, and $A=6r^2$, so $r^2=A/6$. Since $r^3=V$, we have $r^6=V^2=(A/6)^3$, So $216V^2=A^3$. Differentiate with respect to $t$. We get $$(216)(2V)\frac{dV}{dt}=3A^2\frac{dA}{dt}.$$ When $V=216$, $A=216$. Now we can use the above equation to find $\frac{dA}{dt}$ when $V=216$.

I prefer the first way the problem was done, but the second fits better into the "find a relationship, then differentiate" pattern.

Comment: The detailed description of what you did was very helpful in locating the problems. It would be very nice if everybody was this thorough in showing work!

  • $\begingroup$ Thanks for the very detailed explanation too ... I am so careless ... In general, do you have any tips on such problems? Suppose I dont have any hints, how do I figure I should express r in terms of h and how do I know how should I formulate the equation (see update) $\endgroup$ – Jiew Meng Oct 12 '11 at 12:08
  • $\begingroup$ @jiewmeng: These are related rates problems. You know how fast one thing is changing. You want to know how fast another is changing. Find a relationship between the quantities, differentiate. You could have expressed $V$ in terms of $r$ too, find $\frac{dr}{dt}$, and since $h=3r$, $\frac{dh}{dt}=3\frac{dr}{dt}$. After a while related rates problems will all look the same to you, easy. $\endgroup$ – André Nicolas Oct 12 '11 at 13:04
  • $\begingroup$ @jiewmeng: Cube problem analyzed, a couple of ways. $\endgroup$ – André Nicolas Oct 12 '11 at 13:38

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