More than one rate

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Here we work abstract related rates problems.

Suppose we have two, variables $x$ and $y$ , which are both changing with respect to time. A related rates problem is a problem where we know one rate, say $\frac {dx}{dt}$ , at a given instant, and wish to find the other (the unknown rate is ”related” to the known rate).

Here, the chain rule is key: If $y$ is written in terms of $x$ , and we are given $\frac {dx}{dt}$ , then it is easy to find $\frac {dy}{dt}$ using the chain rule: $y=y(x)$ $\frac {dy}{dt}=\frac {dy}{dx}\cdot \frac {dx}{dt}.$ In many cases, particularly the interesting ones, our functions will be related in some other way. Nevertheless, in each case we’ll use the chain rule to help us find the desired rate. In this section, we will work several abstract examples in order to emphasize the mathematical concepts involved. In each of the examples below, we will follow essentially the same plan of attack:

Introduce variables, identify the given rate and the unknown rate.: Assign a variable to each quantity that changes in time.
Draw a picture.: If possible, draw a schematic picture with all the relevant information.
Find equations.: Write equations that relate all relevant variables.
Differentiate with respect to t.: Here we will often use implicit differentiation and obtain an equation that relates the given rate and the unknown rate.
Evaluate and solve.: Evaluate each quantity at the relevant instant and solve for the unknown rate.

Formulas

In order to relate several variables we can use known formulas.

In our next example we consider an expanding circle and use the formulas for perimeter and area of a circle.

Imagine an expanding circle. If we know that the perimeter is expanding at a rate of $4$ m/s, at what rate is the area changing when the radius is $3$ meters?

First, we introduce the variables $P$ , $r$ , and $A$ , denoting the perimeter, the radius, and the area of the circle, in that order.

We identify the given rate, $\frac {dP}{dt}=4$ m/s, and the unknown rate, $\frac {dA}{dt}$ , at the moment when $r=3$ .

Next, we draw a picture.

Now we have to find equations that relate the variables $P$ , $r$ , and $A$ .

Naturally, we think of formulas for perimeter and area of a circle $P = 2\cdot \pi \cdot \answer [given]{r} \qquad \text {and}\qquad A = \pi \cdot \answer [given]{r^{2}}.$ We know that the perimeter of the circle is expanding. This implies that both the radius and the area of the circle are changing in time, too. So, we note that $P$ , $r$ , and $A$ are functions of time $P(t) = 2\cdot \pi \cdot r(t) \qquad \text {and}\qquad A(t) = \pi \cdot r(t)^2.$ We differentiate both sides of each equation with respect to time, $t$ . Using implicit differentiation, we get that

$\underbrace { \dfrac {d}{dt}P(t)}_{given\hspace {0.05in} rate} = 2\cdot \pi \cdot \dfrac {d}{dt}r(t) \qquad \text {and}\qquad \underbrace { \dfrac {d}{dt}A(t)}_{unknown \hspace {0.05in}rate} = 2\cdot \pi \cdot r(t) \cdot \dfrac {d}{dt}r(t).$

These two equations hold on some time interval. In particular, they are both true at an instant when $r=3$ .

We know that at that instant $\dfrac {d}{dt}P(t) = \answer [given]{4}$ and $r = \answer [given]{3}$ .

We now evaluate the two equations at the instant when $r=3$ .

$4 = 2\cdot \pi \cdot \dfrac {d}{dt}r(t) \qquad \text {and}\qquad \dfrac {d}{dt}A(t) = 2\cdot \pi \cdot 3 \cdot \dfrac {d}{dt}r(t)$

From the first equation we get that

$\begin{align*} \frac{d}{dt}r(t)&= 2/\pi \hspace{0.1in} m/s, \hspace{0.1in} when \hspace{0.1in} r=3. \end{align*}$

Using this result and the equation on the right, we get, at the instant when $r=3$ ,

$\begin{align*} \dfrac{d}{dt}A(t) &= 2\cdot \pi\cdot 3 \cdot 2/\pi\\ &=\answer[given]{12}\hspace{0.1in} m^2/s. \end{align*}$ Hence, the area is expanding at a rate of $12$ $m^2/s$ at the instant when $r=3$ m.

Right triangles

In our next example, we consider an expanding right triangle and use the Pythagorean Theorem to relate relevant variables.

Imagine an expanding right triangle. If one leg has a fixed length of $3$ m, one leg is increasing with a rate of $2$ m/s, and the hypotenuse is expanding to accommodate the expanding leg, at what rate is the hypotenuse expanding when both legs are $3$ m long?

First, we introduce the variables $a$ , $b$ , and $c$ , denoting the fixed length, the length of a leg that is increasing, and the length of the hypotenuse, in that order. Then, we identify the given rate $\frac {db}{dt}=2$ m/s and the unknown rate $\frac {dc}{dt}$ , when $b=3$ . Now, we draw a picture.

Next, we find equations that relate relevant variables. Here we use the Pythagorean Theorem. $c^2 = a^2 + b^2$ Note that $a=3$ is a constant, and $c$ and $b$ are functions of time, $t$ . We, then, differentiate both sides of the equation with respect to $t$ , using implicit differentiation

$2\cdot c(t)\cdot \dfrac {d}{dt}c(t) = 2\cdot b\cdot \dfrac {d}{dt}b(t).$ Now, we evaluate all the quantities at the instant when $b=3$ , noting that $\dfrac {db}{dt} = 2$ and that $b = 3$ . Therefore, $2\cdot c\cdot \dfrac {dc}{dt}= \answer [given]{12},\hspace {0.08in} when\hspace {0.08in}b=3.$ We need to compute $c$ , the length of the hypotenuse at the instant when $b=3$ . Here we use the Pythagorean Theorem

$\begin{align*} c^2 &= 3^2 + 3^2\\ &=\answer[given]{18}, \end{align*}$ So, we see that $c = 3\sqrt {2}$ , when $b=3$ .
Then, we solve for the rate at the moment when $b=3$ ,
$\begin{align*} 6\sqrt{2}\cdot \dfrac{dc}{dt} &= 12 \\ \dfrac{dc}{dt} &= \sqrt{2}. \end{align*}$ Therefore, the hypotenuse is growing at a rate of $\answer [given]{\sqrt {2}}$ m/s when both legs are $3m$ long.

Angular rates

We can also investigate problems involving angular rates.

Imagine an expanding right triangle. If one leg has a fixed length of $3$ m, one leg is increasing with a rate of $2$ m/s, and the hypotenuse is expanding to accommodate the expanding leg, at what rate is the angle opposite the fixed leg changing when both legs are $3$ m long?

First, we introduce the variables $a$ , $b$ , $c$ , as in the previous example, and $\theta$ , denoting the angle opposite the leg with fixed length.

We identify the given rate $\frac {db}{dt}=2$ m/s and the unknown rate $\frac {d\theta }{dt}$ , when $b=3$ .

Next, we draw a picture.

We now find equations that combine relevant variables. Here we note that $\tan (\theta ) = \frac {3}{\answer [given]{b}}.$ Since $\theta$ and $b$ are functions of time, we differentiate both sides of the equation using implicit differentiation

$\sec ^2(\theta )\frac {d\theta }{dt} = \frac {-3}{b^2}\cdot \frac {db}{dt}.$ Now, we evaluate all the quantities at the instant when $b=3$ , keeping in mind that $\frac {db}{dt} = 2$ and $b = 3$ , $\sec ^2(\theta )\cdot \dfrac {d\theta }{dt}= \frac {-3}{3^2}\cdot 2\\ = \frac {-2}{3} .$ It is clear that we still need to compute $\sec ^2(\theta )$ at the moment when $b=3$ . A picture will help.

Since we know the lengths of both legs in the right triangle, it is easier to compute

$\tan (\theta )$ than $\sec (\theta )$ . Recall the trig identity: $\sec ^2(\theta )=1+ \tan ^2(\theta )$ .

$\begin{align*} \sec^2(\theta)&= 1+ \tan^2(\theta)\\ &= 1+\Bigl(\frac{3}{3}\Bigr)^2\\ &= \answer[given]{2} \end{align*}$ Now we solve for the unknown rate, when $b=3$ , $\begin{align*} 2\cdot \dfrac{d\theta}{dt} &= \frac{-2}{3}\\ \dfrac{d\theta}{dt}&= \frac{-1}{3} \end{align*}$ So, when $b=3$ , the angle is changing at $\answer [given]{-1/3}$ radians per second.

Similar triangles

Finally, facts about similar triangles are often useful when solving related rates problems.

The figure shows two right triangles that share an (acute) angle.

If $x$ is growing from the vertex with a rate of $3$ m/s, what rate is the area of the smaller triangle changing when $x = 5$ m?

First, we introduce the variables $h$ and $A$ , denoting the height and the area of the smaller triangle, in that order. Next, we identify the given rate $\frac {dx}{dt}=3$ m/s and the unknown rate $\frac {dA}{dt}$ , when $x=5$ . Despite the fact that a nice picture is given, we should do what we always do and draw a picture. Note, we ad information to the picture:

Next, we find equations that combine relevant variables. In this case there are two. The first is the formula for the area of a triangle: $A = \frac {1}{2} \cdot x \cdot h$ The second uses the fact that the larger triangle is similar to the smaller triangle, meaning that the ratios between the corresponding sides in both triangles are equal, $\frac {h}{x} = \frac {3}{\answer [given]{6}}\qquad \text {so}\qquad h = \frac {1}{\answer [given]{2}}\cdot x$ At this point we could differentiate both equations, but we choose a simpler path and express $A$ in terms of $x$

$\begin{align*} A& = \frac{1}{2} \cdot x\cdot\underbrace{ \frac{1}{2}\cdot x}_{h} . \end{align*}$

Therefore, $A=\frac {1}{4}\cdot x^2.$ Since $A$ and $x$ are both functions of time, we differentiate both sides of this equation with respect to $t$

$\frac {dA}{dt}=\frac {1}{4}\cdot 2\cdot x\cdot \frac {dx}{dt}.$ In particular, when $x=5$ , $\frac {dA}{dt}=\frac {1}{4}\cdot 2\cdot 5\cdot 3=\answer {\frac {15}{2}}.$

Hence, the area is changing at a rate of $\frac {15}{2}$ $\text {m}^2/\text {s}$ when $x=5$ m.