# Related Rates

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Lesson Priority: Normal

- Understand the problem at hand and what related rates is trying to accomplish
- Learn the general approach to related rates questions
- Practice the half dozen or so very very common problem types teachers use for exams

When observing an object moving in two dimensions, such as two trains moving away from one another, or a ladder falling down a wall, we can determine a relationship between the speeds of the objects in the system with some Calculus. This topic is many students' enemy, but while related rates problems are admittedly detailed, Mister Math has the step by step way to make these happen.

Practice problems and worksheet coming soon!

## Multiple Movement Measures

Related Rates refers to calculus problems in which two or more related measurements are changing over time (usually distance, area, or volume). The rates at which the measurements are changing are represented as derivatives, and we'll be asked to solve for a number of unknowns. There are a handful of common question types, but any question can have unique aspects.There's no denying the truth - students collectively tend to dislike related rates problems at first. Part of the stigma is that these are word problems, and the other part is the challenge of setting up diagrams ourselves in conjunction with using in-context implicit differentiation.## Related Rates - The Fail-safe Method

Though very problem is different, this approach will always work.- 1. Draw and Label Everything
- 2. Write a relationship equation
- 3. Differentiate
- 4. Plug-in / Solve

- Stick to lines and points when you can - don't actually try to draw the objects.
- Don't make the common mistake of labelling your figure with specific moment information - if a measurement is changing over time, it needs a variable label, not a number.
- Generally speaking, I don't write rates of change on the figure, but rather off to the side. Keep your diagram as clean as possible!

## Common Problem Types

We already saw the moving vehicle problem, but here are a few other frequent fliers.Ladder ProblemsThese problems involve a ladder sliding up or down a wall, and are usually simpler problems because the ladder is a fixed length and the relationship we will use is the Pythagorean one.^{3}/hr. As the pile grows, it retains its shape. Find the rate at which the height is changing at the instant that the radius is $5$ meters.$\blacktriangleright$ Write down what we know and what we want. Let's pick $V$ to represent the volume, since it is not directly part of the figure.$$\mathrm{Know:} \;\; \frac{dV}{dt} = 1.6$$$$\mathrm{Want:} \;\; \frac{dh}{dt} \;\; \mathrm{when} \;\; r=5$$Here we will see the need to use two relationships. Intuitively, we start with the one that relates what we have and what we want - the volume of a cone.$$V = \frac{\pi r^2 h}{3}$$Let's differentiate this relationship using the product rule:$$\frac{dV}{dt} = \frac{\pi}{3} \, \left[ r^2 \cdot \frac{dh}{dt} + 2r \cdot \frac{dr}{dt} \cdot h \right]$$We know $dV/dt$, we want to solve for $dh/dt$, but we weren't given any information about $dr/dt$.Instead, we have to use a second relationship specific to this problem that was described initially:$$h = 4r$$(as stated, the height is equal to four times the radius length, and the pile retains its shape over time)We can glean a relationship between $dr/dt$ and $dh/dt$ by taking the derivative of this relationship.$$\frac{dh}{dt} = 4 \, \frac{dr}{dt}$$or$$\frac{1}{4} \, \frac{dh}{dt} = \frac{dr}{dt}$$Now in our volume derivative equation, we can replace $dr/dt$.$$\frac{dV}{dt} = \frac{\pi}{3} \, \left[ r^2 \cdot \frac{dh}{dt} + 2r \cdot \frac{dr}{dt} \cdot h \right]$$$$\longrightarrow$$$$\frac{dV}{dt} = \frac{\pi}{3} \, \left[ r^2 \cdot \frac{dh}{dt} + 2r \cdot \frac{1}{4} \cdot \frac{dh}{dt} \cdot h \right]$$Finally, we can calculate the rate we seek using the specifics of the instant in question. When $r$ is $5$, we know $h$ must be $20$. Along with the other given info, we have$$1.6 = \frac{\pi \cdot 5^2}{3} \cdot \frac{dh}{dt} + \frac{2(5)\cdot 20}{4} \cdot \frac{dh}{dt}$$$$\frac{dh}{dt} = \frac{1.6}{\frac{25 \pi}{3} + 50} = \frac{4.8}{25 \pi + 150}$$$$\frac{dh}{dt} \approx 0.021 \mathrm{m/hr}$$If you show your work, you probably don't need to try and rationalize or clean up that second-to-last step. Whenever there is $\pi$ in the denominator, it's usually acceptable to turn the exact answer into a decimal approximation. $\blacksquare$

- the total length of the light beam (hypotenuse)
- the length of the light beam between the man and the ground
- the distance between the post and the man
- the distance between the post and the tip of the shadow
- the distance between the man and the tip of the shadow

## Mr. Math Makes It Mean

Related Rates InceptionOne way in which questions become decidedly more mean is to be asked to find a rate that depends on another unknown rate, which you must first find.^{2}/ sec at the moment that the top of the ladder is $6$ ft from the floor. $\blacksquare$

## Put It To The Test

^{2}/ min. $\blacksquare$

^{2}/ min (make sure you write down the units!). $\blacksquare$

^{3}/ sec. How fast is the surface area of the balloon increasing at the moment that the diameter of the balloon is $20$ cm?

^{3}).The volume of a sphere is given by the formula$$V = \frac{4\pi r^3}{3}$$While the given information did not give us the volume, it did give us the rate of change of the volume, which is $dV/dt$. Let's differentiate the volume formula:$$\frac{dV}{dt} = \frac{12 \pi r^2 }{3} \cdot \frac{dr}{dt}$$$$\frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt}$$At the moment at which we are asked to examine, the radius is $10$ cm. Careful not to incorrectly use the given diameter length of $20$ cm in place of the radius (a common trick teachers use)! Now we can solve for $dr/dt$.$$50 = 4 \pi (10)^2 \cdot \frac{dr}{dt}$$$$\frac{dr}{dt} = \frac{1}{8 \pi}$$Now that we know how fast the radius is changing at the moment we want to know about, we can use that information to answer the question that the problem is asking us.To find out how fast the surface area is changing, we'll start with the formula for the surface area of a sphere.$$S = 4 \pi r^2$$Let's differentiate:$$\frac{dS}{dt} = 8 \pi r \cdot \frac{dr}{dt}$$Because we used the given information of the problem to discover the value of $dr/dt$ at the moment that the diameter is $20$ cm, we can finish the problem quickly.$$\frac{dS}{dt} = 8 \pi (10) \cdot \frac{1}{8 \pi}$$$$\frac{dS}{dt} = 10$$Therefore, at the moment that this inflating spherical balloon has a diameter of $20$ cm, the surface area is changing at a rate of $10$ cm

^{2}/ sec. $\blacksquare$

- Understand the premise of Related Rates problems
- Become a master of figure drawing and labelling
- Know the four step process to handle any related rates problem
- Become very familiar with the usual kinds of problems you'll encounter

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