More than one rate

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}} \DeclareMathOperator {\proj }{\vec {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\boldsymbol {\l }}} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

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 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 $\dd [x]{t}$ , then it is easy to find $\dd [y]{t}$ using the chain rule: $y=y(x)$ $\dd [y]{t}=y'(x(t))\cdot x'(t).$ In many cases, particularly the interesting ones, our functions will be related in some other way. Nevertheless, in each case we’ll use the power of the chain rule to help us find the desired rate. In this section, we will work several abstract examples, so we can emphasize the mathematical concepts involved. In each of the examples below, we will follow essentially the same plan of attack:

Introduce variables.: Assign a variable to each quantity that changes in time.
Identify the given and unknown rates.
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 time t.: Here we will often use implicit differentiation and obtain an equation that relates the given rate and the unknown rate.
Evaluate.: Evaluate each quantity at the relevant moment.
Solve.: Solve for the unknown rate at that moment.

Formulas

One way to combine several variables is with a known formula.

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

First, we introduce the variables $P$ , $r$ , and $A$ . Let $P$ denote the perimeter of the circle, let $r$ denote the radius of the circle, and let $A$ denote the area of the circle. Then, we identify the given rate $\frac {dP}{dt}=4$ m/s and the unknown rate $\Bigl [\frac {dA}{dt}\Bigr ]_{r=3}$ , the rate to be determined. Next, we draw a picture.

Next, we find equations that combine relevant variables. Here we use the common formulas for perimeter and area $P = \answer [given]{2\cdot \pi \cdot r} \qquad \text {and}\qquad A = \answer [given]{\pi \cdot r^2}.$ 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 $A$ , $r$ , and $P$ 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 using implicit differentiation, treating all functions as functions of $t$ . $\dfrac {d}{dt}P(t) = 2\cdot \pi \cdot \dfrac {d}{dt}r(t) \qquad \text {and}\qquad \dfrac {d}{dt}A(t) = 2\cdot \pi \cdot r(t) \cdot \dfrac {d}{dt}r(t).$ Now we evaluate all the quantities at the moment when $r=3$ .

We know that at that moment $\Bigl [ \dfrac {d}{dt}P(t)\Bigr ]_{r=3} = \answer [given]{4}$ and that $r = \answer [given]{3}$ . Hence our equations become $4 = 2\cdot \pi \cdot \Bigl [\dfrac {d}{dt}r(t)\Bigr ]_{r=3} \qquad \text {and}\qquad \Bigl [ \dfrac {d}{dt}A(t)\Bigr ]_{r=3} = 2\cdot \pi \cdot 3 \cdot \Bigl [\dfrac {d}{dt}r(t) \Bigl ]_{r=3}.$ We see that

$\begin{align*} 4 &= 2\cdot \pi\cdot \Bigl[\dfrac{d}{dt}r(t)\Bigr]_{r=3}\\ 2/\pi &= \Bigl[\dfrac{d}{dt}r(t)\Bigr]_{r=3}. \end{align*}$ Therefore, $\Bigl [\dfrac {d}{dt}r(t)\Bigr ]_{r=3}= 2/\pi$ m/s. Now we solve for the rate of $A$ at the moment when $r=3$ . $\begin{align*} \Bigl[ \dfrac{d}{dt}A(t)\Bigr]_{r=3} &= 2\cdot \pi\cdot 3 \cdot 2/\pi\\ &=\answer[given]{12}. \end{align*}$ Hence the area is expanding at a rate of $12$ $m^2/s$ at the instant when $r=3$ m.

Right triangles

A common way to combine variables is through facts related to right triangles.

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 $c$ . Let $a$ denote the length of a leg with fixed length, let $b$ denote the length of a leg whose length is increasing, and let $c$ denote the length of the hypotenuse. Next, we identify the given rate $\frac {db}{dt}=2$ m/s and the unknown rate $\Bigl [\frac {dh}{dt}\Bigr ]_{b=3}$ , the rate to be determined. Now, we draw a picture.

Next, we find equations that combine relevant variables. Here we use the Pythagorean Theorem. $c^2 = a^2 + b^2$ We note that $c$ and $b$ are functions of time, and write $c(t)^2 = a^2 + b(t)^2.$ We differentiate both sides of the equation using implicit differentiation, treating all functions as functions of $t$ , note $a$ is constant, $2\cdot c(t)\cdot \dfrac {d}{dt}c(t) = 2\cdot b(t)\cdot \dfrac {d}{dt}b(t).$ Now, we evaluate all the quantities at the moment when $b=3$ . We know that $\Bigl [\dfrac {d}{dt}b(t)\Bigr ]_{b=3} = 2$ and that $b = 3$ $2\cdot [c(t)]_{b=3}\cdot \Bigl [\dfrac {d}{dt}c(t)\Bigr ]_{b=3} = \answer [given]{12}$ However, we still need to know $[c(t)]_{b=3}$ , the length of the hypotenuse at the moment when $b=3$ . Here we use the Pythagorean Theorem,

$\begin{align*} \Bigl([c(t)]_{b=3}\Bigr)^2 &= 3^2 + 3^2\\ &=\answer[given]{18}, \end{align*}$ and so we see that $[c(t)]_{b=3} = 3\sqrt {2}$ .
And we solve for the rate.
$\begin{align*} 6\sqrt{2}\cdot \Bigl[\dfrac{d}{dt}c(t)\Bigr]_{b=3} &= 12 \\ \Bigl[\dfrac{d}{dt}c(t)\Bigr]_{b=3} &= \sqrt{2}. \end{align*}$ Hence $c(t)$ 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 $a$ , $b$ , $c$ , and $\theta$ . Let $a$ denote the length of a leg with fixed length, let $b$ denote the length of a leg whose length is increasing, let $c$ denote the length of the hypotenuse, and let $\theta$ denote the angle opposite the side with fixed length. Next, we identify the given rate $\frac {db}{dt}=2$ m/s and the unknown rate $\Bigl [\frac {d\theta }{dt}\Bigr ]_{b=3}$ , the rate to be determined. Next, we draw a picture.

We now find equations that combine relevant functions. Here we note that $\tan (\theta ) = \answer [given]{\frac {a}{b}}$ Since $\theta$ and $b$ are functions of time, we write $\tan (\theta (t)) = \frac {a}{b(t)}.$ We differentiate both sides of the equation using implicit differentiation, treating all functions as functions of $t$ , note $a$ is constant, $\sec ^2(\theta (t))\theta '(t) = \frac {-a\cdot b'(t)}{b(t)^2}.$ Now we evaluate all the quantities at the moment when $b=3$ . We know that $a=3$ , $b'(t) = 2$ , and that $b = 3$

$\begin{align*} \Bigl[\sec^2(\theta(t))\Bigr]_{b=3}\cdot \Bigl[\dfrac{d}{dt}\theta(t)\Bigr]_{b=3} &= \frac{-3\cdot 2}{3^2}\\ &= \frac{-6}{9}\\ &= \frac{-2}{3}. \end{align*}$ However, we still need to compute $\Bigl [\sec ^2(\theta (t))\Bigr ]_{b=3}$ . Here we use the Pythagorean Theorem, $\begin{align*} [c^2(t)]_{b=3} &= 3^2 + 3^2\\ &=\answer[given]{18}, \end{align*}$ and so we see that $[c(t)]_{b=3} = 3\sqrt {2}$ . Now $\begin{align*} \Bigl[\sec^2(\theta(t))\Bigr]_{b=3} &= \frac{\mathrm{hypotenuse}^2}{\mathrm{adjacent}^2}\\ &= \frac{\left(3\sqrt{2}\right)^2}{3^2}\\ &= \answer[given]{2}. \end{align*}$ Now we solve. $\begin{align*} \Bigl[\sec^2(\theta(t))\Bigr]_{b=3}\cdot \Bigl[\dfrac{d}{dt}\theta(t)\Bigr]_{b=3} &= \frac{-2}{3}\\ 2\cdot \Bigl[\dfrac{d}{dt}\theta(t)\Bigr]_{b=3} &= \frac{-2}{3}\\ \Bigl[\dfrac{d}{dt}\theta(t)\Bigr]_{b=3} &= \frac{-1}{3}. \end{align*}$ So when $a=b=3$ , 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.

Imagine two right triangles that share an 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 $h$ , the height of the smaller triangle, and $A$ , the area of the smaller triangle. Next, we identify the given rate $\frac {dx}{dt}=3$ m/s and the unknown rate $\Bigl [\frac {dA}{dt}\Bigr ]_{x=5}$ , the rate to be determined. 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: $A = \answer [given]{(1/2) \cdot x \cdot h}$ 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, $\frac {x}{h} = \answer [given]{\frac {6}{3}}\qquad \text {so}\qquad x = \answer [given]{2}\cdot h$ Since $A$ , $x$ , and $h$ are functions of time, we write $A(t) = (1/2) \cdot x(t) \cdot h(t) \qquad \text {and}\qquad x(t) = 2\cdot h(t).$ We now differentiate both sides of each equation using implicit differentiation, treating all functions as functions of $t$ ,

$\begin{align*} \dfrac{d}{dt}A(t) &= (1/2) \cdot\dfrac{d}{dt}x(t) \cdot h(t) + (1/2) \cdot x(t) \cdot \dfrac{d}{dt}h(t),\\ \dfrac{d}{dt}x(t) &= 2\cdot\dfrac{d}{dt}h(t) \end{align*}$ Now we evaluate all the quantities at the moment when $x=5$ m. $5 = 2\cdot [h(t)]_{x=5}\qquad \text {and}\qquad 3= 2\cdot \Bigl [\dfrac {d}{dt}h(t)\Bigr ]_{x=5}$ we see that $[h(t)]_{x=5} = 5/2$ and $\Bigl [\dfrac {d}{dt}h(t)\Bigr ]_{x=5} = 3/2$ .
Now we solve for the rate. $\begin{align*} \Bigl[\dfrac{d}{dt}A(t)\Bigr]_{x=5} &= (1/2) \cdot 3 \cdot (5/2) + (1/2) \cdot 5 \cdot (3/2)\\ &= 15/4 + 15/4\\ &= \answer[given]{15/2}. \end{align*}$ Hence, the area is changing at a rate of $15/2$ $\text {m}^2/\text {s}$ when $x=5$ m.