So the step where you implicit derive the geometric formulas for time seem to be giving me the most trouble.

How would this be done for volume of a cylinder $$(v=pi^2h)$$ and area of a triangle $$(1/2bh)$$

For the latter, I gave it a try and got $$(b*db/dt*h*dh/dt)/0$$$. Considering the derivative of a constant (2) is zero, the equation is undefined and I don't see how this could be the right equation

For the former equation, the question asks: Water is being pumped into a vertical cylinder of radius 5 meters and height 20 meters at a rate of 3 meters3/min. How fast is the water level rising when the cylinder is half full?

For the later, the question asks: A right triangle has one leg of 7 cm. How fast is its area changing at the instant that the other leg has length 10 cm and is decreasing at 2 cm per second?


1 Answer 1


For the first question, your formula is wrong. The volume is given by $V = \pi r^2 h$. You omitted the radius.

Radius is constant, the height (of the water level) is the changing variable.

So the derivative with respect to time is:

$\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$

For the second question, $A = \frac{1}{2}b\cdot h$.

You're keeping one leg (call that $h$) constant.

So the time-derivative is:

$\frac{dA}{dt} = \frac{1}{2}b\frac{dh}{dt}$.

In your post, you also misapplied a rule (not sure if you thought you were applying Chain or Product rule, but the form is still wrong).

  • $\begingroup$ I see. Could you explain the second one a little more? How did you know to take the height rather than the base from the problem? Why isn't the derivative of (1/2), 0? $\endgroup$
    – aero26
    Nov 7, 2014 at 6:59
  • $\begingroup$ Doesn't matter which is called the height or the base, as long as it's one of the two perpendicular sides in a right triangle (technical name is "cathetus"). You're just assigning one (the constant one) to be $b$, you could just as easily have done it the other way around. The only way the problem could be changed (made more difficult) is if they told you the changing side was the hypotenuse (which is quite different from either of the catheti). (more to follow) $\endgroup$
    – Deepak
    Nov 7, 2014 at 7:05
  • $\begingroup$ (cont'd) With regard to the derivative of the constant, it works differently depending on whether the constant is added or multiplied to an expression. The derivative of $f(x) + C$ is $f'(x)$ (i.e. the constant vanishes when it's added to an expression) but the derivative of $Cf(x)$ is $Cf'(x)$ (i.e the constant remains the same when it's multiplied to an expression). I've been writing very loosely here (without rigour), but that's just to make things simple - hopefully, you understand exactly what I mean. $\endgroup$
    – Deepak
    Nov 7, 2014 at 7:07

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