At What Rate Is The Exposed Surface Area Of The Water Changing When The Radius Is 2 Feet
A student recently wrote to ask if we'd aid solve a common related rates problem almost water draining from a cylindrical tank. The trouble was something similar this:
Cylinder Drains Water.
A cylinder filled with water has a 3.0-pes radius and 10-foot height. It is tuckered such that the depth of the water is decreasing at 0.1 feet per second. How fast is the water draining from the tank?
Remember that related rates problems ever requite y'all the rate of one quantity that's changing, and ask yous to find the rate of something else that changes every bit a upshot. Hither, the problem tells yous that the water level is falling at 0.1 ft/s, and asks for the rate at which the book of water in the tank is decreasing. That is, if we call the volume of water in the tank at any moment V, then what is dV/dt?
In an before post, we adult a 4-Footstep Strategy to solve almost any Related Rates problem.
Problem SOLVING STRATEGY: Related Rates
Hide/Show Strategy
- Describe a picture of the concrete situation.
Don't stare at a bare piece of paper; instead, sketch the situation for yourself. Really. - Write an equation that relates the quantities of interest.
- Be sure to label as a variable whatsoever value that changes as the state of affairs progresses; don't substitute a number for it yet.
- To develop your equation, you volition probably use:
- a simple geometric fact (like the relation between a circle's area and its radius, or the relation between the volume of a cone and its base-radius and height);
- a trigonometric function (like $\tan{\theta}$ = opposite/next);
- the Pythagorean theorem;
- similar triangles.
Most frequently (> 80% of the time) you lot will employ the Pythagorean theorem or similar triangles.
- Take the derivative with respect to time of both sides of your equation. Remember the Chain Rule.
- Solve for the quantity you lot're after.
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Let's use the strategy to solve this problem.
one. Draw a flick of the physical state of affairs.
See the figure. Let's call the superlative (or depth) of the water at any given moment y, as shown.
When a quantity is decreasing, we have to make the rate negative.
We are told that the h2o level in the cup is decreasing at the charge per unit of $0.one\, \tfrac{\text{ft}}{\text{s}}$, then $\dfrac{dy}{dt} = -0.1\, \tfrac{\text{ft}}{\text{due south}}$. Remember that we have to insert that negative sign "by paw" since the water's height is decreasing.
two. Write an equation that relates the quantities of interest.
A. Be sure to characterization equally a variable any value that changes every bit the state of affairs progresses; don't substitute a number for information technology yet.
The elevation of the water changes as time passes, and so we're calling that the variable y.
B. To develop your equation, you volition probably use . . . a unproblematic geometric fact.
The book Five of whatsoever cylinder is its round cross-sectional surface area $\left(\pi r^2 \right)$ times its peak. Here, at any moment the water's meridian is y, and so the volume of water in the cylinder is:
$$V = \pi r^2y$$
3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
Note that the cylinder's radius, r, is constant (r = 3.0 ft), so we'll treat it as a abiding when nosotros take the derivative. By contrast, the water'due south summit y is non constant; instead it changes, and indeed, it changes at the rate dy/dt.
\brainstorm{marshal*}
\frac{dV}{dt} &= \frac{d}{dt}\left(\pi r^2 y \right) \\ \\
&= \pi r^2 \frac{d}{dt}(y) \\ \\
&= \pi r^2 \frac{dy}{dt} \\ \\
\end{align*}
iv. Solve for the quantity you're after.
At this signal we're just substituting values: the problem told us $r = 3.0 \, \text{ft}$ and $\dfrac{dy}{dt} = -0.i\, \tfrac{\text{ft}}{\text{s}}$, and we're looking for dV/dt. Starting from our last expression above:
\begin{align*}
\frac{dV}{dt} &= \pi r^2 \frac{dy}{dt} \\ \\
&= \pi (iii.0\, \text{ft})^ii \left(-0.ane\, \tfrac{\text{ft}}{\text{due south}}\right) \\ \\
&= -2.eight \, \tfrac{\text{ft}^iii}{\text{s}} \quad \cmark
\end{align*}
That is, h2o is draining from the tank at the rate $\dfrac{dV}{dt} = -2.8\, \tfrac{\text{ft}^3}{\text{s}}$. The negative value indicates that the h2o'south volume in the tank Five is decreasing, which is correct.
A related, harder problem that's common on exams
Another very common Related Rates trouble examines water draining from a cone, instead of from a cylinder. While the idea is very much the same, that problem is a little more challenging because of a sub-trouble required to deal with the cone's geometry. You can acquire how to solve that in this web log post.
Time to exercise
You must practice for yourself, pencil in your mitt, before your exam.
Reading our solution may be a not bad showtime to larn how to solve Related Rates problems, but to be fix for your examination you lot need to exercise for yourself, pencil in your hand. That'southward when y'all'll get stuck and make mistakes and do all the other things people exercise when they're learning something new. . . and you want to exercise that before your examination! We take lots of problems for you lot to effort, each with a complete step-past-step solution.
For more than example problems with complete solutions, please visit our gratuitous Related Rates folio!
Over to you:
- What tips practise you take to share almost solving Related Rates issues?
- What questions do you accept?
We'd love to hear from you. Delight comment below!
Source: https://www.matheno.com/blog/related-rates-problem-cylinder-drains-water/
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