If there is a cylinder with a base radius of 70cm, and water is being poured into it at 10 liters per minute, how fast is the water level rising?
I've gotten myself horribly confused as to how to do this; can anyone provide any insight?
Thanks!
[Math] Related Rates – Calculus – Filling a cylinder
calculus
Related Solutions
Think about what is happening. The water (volume) is being poured in at a constant rate. This relates to how the water level ($h$) changes and how the width of the water in the tank at that level ($r$) changes. Further, $r$ and $h$ are related.
The volume of a cone is
$$V = \frac13 \pi \, r^2 \, h$$
How is $h$ related to $r$? You know that, at the top, the radius is $2$ and $h=4$. Because this is a cone, we can say that $r = h/2$ at all levels of the cone. Thus,
$$V(h) = \frac{1}{12} \pi \, h^3$$
We may then differentiate with respect to time; use the chain rule here
$$\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$$
You are given $dV/dt$ and the height $h$ at which to evaluate; solve for $dh/dt$.
Step 0, draw a diagram. Try to imagine these objects moving around in your head and immediately observe any relationships between these variables. If there's an unknown rate, label it $\frac{d?}{d??}$. Most of the time, these problems are not hard to solve (calculus wise), it's all visualization.
Recall, the rate in the denominator, $d??$ is the "with respect to" portion.
Step 1, list all of our knowns:
1.$$V = \frac{1}{3}\pi r^2h$$ 2.$$\frac{dV}{dt} = \frac{-50m^3}{min}$$ 3.$$h = 6$$ 4.$$h_2 = 5$$ 5.$$r = 45$$
Step 2, figure out what we need to solve for, this is your $\frac{d?}{d??}$ In this case, we're looking at the rate of change in height with respect to our radius. Always ask yourself: what is this problem asking me to solve? Write it in terms of $\frac{d?}{d??}$, which in our case is $\frac{dh}{dt}$.
Step 2a, imagine the rate of change you need to solve for. The question asked "at what rate is the water level falling at a height of 5m [in the conic]". So, imagine in your head the water sliding down the conic. Ask yourself, "what's changing?" In this case, the water is sliding down the side of the conic, and as it does so, the radius (and thereby the volume) is changing!
Step 3, identify relationships between variables, keeping in mind our step 2 and step 2a. In this case, it is the relationship between $r$ and $h$ that is so important, so we're going to setup a ratio between $r$ and $h$:
$$\frac{r}{h} = \frac{45}{6}$$ $$ \bf{r = \frac{45h}{6}}$$
Step 4, substitute and simplify as much as reasonably possible.
$$V = \frac{1}{3}\pi r^2h$$ $$V = \frac{1}{3}\pi \left(\bf{\frac{45h}{6}}\right)^2 h$$ $$V = \frac{1}{3}\pi \left(\bf{\frac{15h}{2}}\right)^2 h$$
Step 5, finally, you get to differentiate; however, as this is probably a homework question, I'm going to leave that part to you. One thing to remember is that you don't start plugging numbers in, really until you're almost at the end. Notice here you're subbing in your final $h_2$ value. You're going to get something like:
$$\frac{dV}{dt} = \frac{some number}{some number} \pi \bf{h_2}^2 \bf{\frac{dh}{dt}} $$
Simplify that down and algebraically manipulate it until you get:
$$ some number = \frac{dh}{dt}$$ and that's your answer.
So, how does all this work? (Read: what am I doing?)
So, as you define these relationships that govern this changing thing, what you're really doing is setting up the problem and differentiating to find the rate of change for that problem for some arbitrary instance. After you find the derivative, you substitute in those specific values, in our case your $h_2$, and literally freezing the rate at an exact moment in time to get the rate at which the quantity is changing.
Let me know if this helps.
Best Answer
Hints: You have a cylinder with height $h$ and radius of the base $r$ and volume $V$. Then $$ V = \pi r^2h = \pi (7dm)^2h. $$ (Using this the volume will be in liters since $1$ liter is $(10$cm$)^3$ and $10$cm$=1$dm (decimeter). The height is now measured in decimeters).
You know that $$ \frac{dV}{dt} = 10 $$ and you want to find $$\frac{dh}{dt}.$$ What you can do is to use the colume equation above and use implicit differentiation to relate $\frac{dV}{dt}$ and $\frac{dh}{dt}$.