I was working through my calculus textbook's practice problems when I came across a problem I couldn't figure out. I really suck at rate of change word problems and this one stumped me. Unfortunately, the textbook only gives answers to even-numbered problems and this is an odd question. Anyway it goes something like this:
An empty oil container is 10 meters long. A cross-section of the
container is in the shape of an isosceles trapezoid that is 30cm wide
at the bottom and 80cm wide at the top and has a height of 50cm. The
container is filled with oil at a rate of 2 meters cubed per minute.
How fast is the oil level rising when it is 30cm deep?
I would give my work but I have little to no idea if it's right and I'm afraid I would be just wasting my time as it's probably wrong. How do you do problems like these? I feel like I was never properly taught how and I would like to have a solid understanding by the time I'm actually being tested on this material.
Best Answer
Refer to the figure:
The volume of the container filled with oil is: $$V=\frac{(0.3+2x)+0.3}{2}\cdot h\cdot 10.$$
From the similarity of the two triangles on the left we get: $$\frac{x}{0.25}=\frac{h}{0.5} \Rightarrow x=\frac{h}{2}.$$
Substitute this into the volume formula: $$V=\frac{(0.3+2\cdot \frac{h}{2})+0.3}{2}\cdot h\cdot 10=\frac{(h+0.6)h}{2}\cdot 10=5h^2+3h.$$ Take the derivative of the volume function with respect to time: $$\frac{dV}{dt}=\frac{dV}{dh}\cdot \frac{dh}{dt}=2 \ \frac{m^3}{\text{min}} \Rightarrow \\ (10h+3)\cdot \frac{dh}{dt}=2 \Rightarrow \\ (10\cdot 0.3+3)\cdot \frac{dh}{dt}=2 \Rightarrow \\ \frac{dh}{dt}=\frac{2}{6}=\frac13.$$