Applied related rates

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We solve related rates problems in context.

Now we are ready to solve related rates problems in context. Just as before, we are going to follow essentially the same plan of attack in each problem.

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

A hand-tossed pizza crust starts off as a ball of dough with a volume of $400\pi \, \text {cm}^3$ . First, the cook stretches the dough to the shape of a cylinder of radius $12$ cm. Next the cook tosses the dough.

If during tossing, the dough maintains the shape of a cylinder and the radius is increasing at a rate of $15$ cm/min, how fast is its thickness changing when the radius is $20$ cm?

First, we introduce the variables $V$ , $r$ , and $h$ . Let $V$ denote the volume of the pizza dough, let $r$ denote the radius of the pizza, and let $h$ be the thickness of the pizza. Then, we identify the given rate $\frac {dr}{dt}=15$ cm/min and the unknown rate $\Bigl [\frac {dr}{dt}\Bigr ]_{r=20}$ , the rate to be determined.

Next, we draw a picture. Here we see a cylinder that represents our pizza dough.

Next we need to find equations. We see that we have $400\pi = \answer [given]{\pi \cdot r^2 \cdot h},$ which immediately simplifies to $400 = r^2 \cdot h.$ Since both $r$ and $h$ are functions of time, we now write $400 = r(t)^2 \cdot h(t).$ Now we differentiate both sides of the equation using implicit differentiation, treating all functions as functions of $t$ , $0 = 2\cdot r(t) \cdot r'(t) \cdot h(t) + r(t)^2 \cdot h'(t).$ Now we’ll evaluate all the quantities at the moment when $r=20$ . But, first, we have to compute $h$ at that moment. Since $400 = 20^2 \cdot h,$ it follows that $h=1$ , when $r=20$ . Now, we are ready to solve.

$0 = 2\cdot 20 \cdot 15 \cdot 1 + 20^2 \cdot \Bigl [\frac {d}{dt}h\Bigr ]_{r=20}$

$\begin{align*} -2\cdot 20 \cdot 15 &= 20^2 \cdot \Bigl[\frac{d}{dt}h\Bigr]_{r=20}\\ \frac{-2\cdot 20 \cdot 15}{20^2} &= \Bigl[\frac{d}{dt}h\Bigr]_{r=20}\\ -1.5 &= \Bigl[\frac{d}{dt}h\Bigr]_{r=20} \end{align*}$ Hence, the thickness of the dough is changing at a rate of $\answer [given]{-1.5}$ cm/min.

Consider a melting snowball. We will assume that the rate at which the snowball is melting is proportional to its surface area. Show that the radius of the snowball is changing at a constant rate.

First, we introduce the variables $V$ , $r$ , and $A$ . Let $V$ denote the volume of the snowball, let $r$ denote the radius of the snowball, and let $A$ be the surface area of the snowball . Then, we identify the given rate $\frac {dV}{dt}$ and the unknown rate $\frac {dr}{dt}$ , the rate to be determined. This problem is a bit unusual, because ”the given rate” is not explicitly given. We will deal with this issue below. Next, we draw a picture.

Now we need to find equations that relate all relevant variables. The equations we’ll use are $V = \answer [given]{(4/3) \cdot \pi \cdot r^3} \qquad \text {and}\qquad A = \answer [given]{4\cdot \pi \cdot r^2}.$ Now the key words are “the rate at which the snowball is melting (the ”given” rate) is proportional to its surface area.” From this we have the following equation:

We need to compute $\frac {dV}{dt}$ . We know $V = \frac {4}{3}\cdot \pi \cdot r^3$ . Since $r$ is a function of $t$ , we write $V(t) = \frac {4}{3}\cdot \pi \cdot r(t)^3.$ So $\frac {dV}{dt} = 4\cdot \pi \cdot r(t)^2 \cdot \frac {dr}{dt}.$ We also know that $\frac {dV}{dt} = k\cdot A(t).$ So, $4\cdot \pi \cdot r(t)^2 \cdot \frac {dr}{dt} = k\cdot 4\cdot \pi \cdot r(t)^2 .$ Therefore, we solve for $\frac {dr}{dt}$ and obtain that $\begin{align*} \frac{dr}{dt} &= k.\\ \end{align*}$ Hence, the radius is changing at a constant rate. Notice, in this example we did not have to evaluate quantities at particular time, because the unknown rate did not depend on time.

Right triangles

A road running north to south crosses a road going east to west at the point $P$ . Cyclist $A$ is riding north along the first road, and cyclist $B$ is riding east along the second road. At a particular time, cyclist $A$ is $3$ kilometers to the north of $P$ and traveling at $20$ km/hr, while cyclist $B$ is $4$ kilometers to the east of $P$ and traveling at $15$ km/hr. How fast is the distance between the two cyclists changing at that time?

First, we introduce the variables $a$ , $b$ , and $c$ . Let $a$ denote the distance of cyclist $A$ from the point $P$ , let $b$ denote denote the distance of cyclist $B$ from the point $P$ , and let $c$ denote the distance between the two cyclists. We identify the given rates $\Bigl [\frac {da}{dt}\Bigr ]_{a=3, b=4}=20$ km/h, $\Bigl [\frac {db}{dt}\Bigr ]_{a=3, b=4}=15$ km/h and the unknown rate $\Bigl [\frac {dc}{dt}\Bigr ]_{a=3, b=4}$ . Now, we draw a picture.

We find equations relating variables $a$ , $b$ , and $c$ . By the Pythagorean Theorem, $c(t)^2=a(t)^2+b(t)^2.$ Now we differentiate both sides of the equation. $2c(t)c'(t)=2a(t)a'(t)+2b(t)b'(t).$ Next, we will evaluate all the quantities in this equation at the time when $a=3$ and $b=4$ . We see that we $c$ needs to be evaluated at that time, too. In order to do that, let’s draw and label the picture at the time when $a=3$ and $b=4$ .

By the the Pythagorean Theorem and the picture above, it follows that $c = \answer [given]{5} km$ Now we evaluate all the quantities at the time when $a=3$ and $b=4$ , and get $2\cdot 5 \cdot \Bigl [\frac {dc}{dt}\Bigr ]_{a=3, b=4} = 2 \cdot 3\cdot 20 + 2 \cdot 4 \cdot 15.$ Solving for $\Bigl [\frac {dc}{dt}\Bigr ]_{a=3, b=4}$ we find that

$\Bigl [\frac {dc}{dt}\Bigr ]_{a=3, b=4} = \answer [given]{24}$ km/hr.

A plane is flying at an altitude of $3$ miles directly away from you at $500$ mph . How fast is the plane’s distance from you increasing at the moment when the plane is flying over a point on the ground $4$ miles from you?

First, we introduce variables $p$ , the distance the plane has traveled after the moment when it flew right above you, and $s$ , the distance between you and the plane. The given rate is $\frac {dp}{dt}=500$ mph and the unknown (related) rate is $\Bigl [\frac {ds}{dt}\Bigr ]_{p=4}$ . Next, we draw a picture.

Next we find equations relating the variables $p$ and $s$ . By the Pythagorean Theorem we know that $p^2+3^2=s^2.$ Since $p$ and $s$ are functions of time, we now differentiate both sides of the equation $2\cdot p(t)\cdot p'(t) = 2\cdot s(t) \cdot s'(t).$ Next we evaluate all the quantities at the moment when $p=4$ miles. This implies that $s$ has to be evaluated at time when $p=4$ . In order to do that, let’s draw the picture at the moment when $p=4$ mi.

By applying the Pythagorean Theorem, we find that $4^2+3^2=s^2$ , so $s=\answer [given]{5}$ . Putting together all the information we get $2(4)(500)=2(5)\Bigl [\frac {ds}{dt}\Bigr ]_{p=4} .$ Finally, we solve: $\Bigl [\frac {ds}{dt}\Bigr ]_{p=4}=\answer [given]{400}$ mph.

Angular rates

A plane is flying at an altitude of $3$ miles directly away from you at $500$ mph . Let $\theta$ be the angle of elevation of the plane, i.e., the angle between the ground and your line of sight to the plane. How fast is the angle $\theta$ decreasing at the moment when the plane is flying over a point on the ground $4$ miles from you?

First, we introduce a variable $p$ , the distance the plane has traveled after the moment when it flew right above you. So, the given rate is $\frac {dp}{dt}=500$ mph and the unknown (related) rate is $\Bigl [\frac {d\theta }{dt}\Bigr ]_{p=4}$ . This rate should be measured in rad/s. Therefore, we have to convert the units of the given rate, mph, into mi/s: $\frac {dp}{dt}=500$ mph $=\frac {500}{60\cdot 60}$ mi/s.
Next, we draw a picture.

Now we find an equation relating variables $p$ and $\theta$ . From the picture we can see that $\tan {\theta }=\frac {3}{p}.$ Since $p$ and $\theta$ are both functions of time, we now differentiate both sides of the equation. We write $\sec ^{2}{\theta }\cdot \theta ' = -\frac {3}{p^2}\cdot p'.$ The above equation holds over some interval of time. In particular, it is true when $p=4$ . Now, we draw and label the picture at the moment when $p=4$ mi.

We now evaluate all the quantities at the moment when $p=4$ . First, we have to compute $\sec ^{2}{\theta }$ at that time. We will remind ourselves of the trig identity $\sec ^2{\theta }=1+\tan ^2{\theta }.$ So, $\Big [\sec ^2{\theta }\Bigr ]_{p=4}=1+\Bigr (\frac {3}{4}\Bigl )^2=\frac {25}{16}.$ Now, by substituting all the known values into the equation above, we get $\frac {25}{16}\cdot \Bigl [\frac {d\theta }{dt}\Bigr ]_{p=4} = -\frac {3}{16}\cdot \frac {500}{60\cdot 60}.$ We solve for $\Bigl [\frac {d\theta }{dt}\Bigr ]_{p=4}$ and get that $\Bigl [\frac {d\theta }{dt}\Bigr ]_{p=4} = -\frac {1}{60} rad/s.$

Similar triangles

It is night. Someone who is $6$ feet tall is walking away from a street light at a rate of $3$ feet per second. The street light is $15$ feet tall. The person casts a shadow on the ground in front of them. How fast is the length of the shadow growing when the person is $7$ feet from the street light?

First, we introduce two variables, $p$ and $s$ . Let $p$ be the distance from the person to the lamp, and let $s$ be the length of the shadow. So, the given rate is $\frac {dp}{dt}=3$ ft/s and the unknown (related) rate is $\Bigl [\frac {ds}{dt}\Bigr ]_{p=7}$ . Then, we draw a picture.

Now we find equations that relate the variables $p$ and $s$ . We use the fact that we have similar triangles to write:

$\begin{align*} \frac{s+p}{\answer[given]{15}} &= \frac{s}{\answer[given]{6}},\\ 6\cdot s + 6 \cdot p &= 15\cdot s,\\ 6\cdot p &=9\cdot s,\\ 2\cdot p &=3 \cdot s. \end{align*}$ Since, $p$ and $s$ are both functions of time, we differentiate both sides of the equation above. We use implicit differentiation and write $2\cdot p'(t) =3 \cdot s'(t)$ At this point we evaluate all the quantities at time when $p=7$ ft $2\cdot 3 = 3 \cdot \Bigl [\frac {ds}{dt}\Bigr ]_{p=7}.$ Now we solve for $\Bigl [\frac {ds}{dt}\Bigr ]_{p=7}$ and get that $\Bigl [\frac {ds}{dt}\Bigr ]_{p=7} = \answer [given]{2}$ , meaning the shadow is growing at a rate of $2$ feet per second when the person is $7$ ft from the lamp.

Water is poured into a conical container at the rate of 10 cm ${}^3$ /s. The cone points directly down, and it has a height of 30 cm and a base radius of 10 cm. How fast is the water level rising when the water is 4 cm deep?

First, we introduce several variables, $V$ , $r$ , and $h$ . Let $V$ denote the volume of the water in the container, let $r$ denote the radius of the circlular surface of the water, and let $h$ be the depth of the water in the container. The given rate is $\frac {dV}{dt}=10$ $cm^3/s$ , and the unknown rate is $\Bigl [\frac {dh}{dt}\Bigr ]_{h=4}$ . Now, we draw a picture.

Note, no attempt was made to draw this picture to scale, rather we want all of the relevant information to be available to the mathematician.

Now we find equations that relate all the variables. Notice that water in the container assumes the shape of the container. In this example, the shape is a cone. Therefore, we use the formula for the volume of a cone $V = \frac {\pi }{3} r^2 h.$ Let’s draw a cross section of the cone.

Notice a big right triangle and a smaller right triangle inside the big one. These two right triangles are similar, because their corresponding angles are equal. Since the ratios of corresponding sides in similar triangles are equal, we get $\frac {r}{h}=\frac {10}{30} \qquad \text {so}\qquad r=\answer [given]{h/3}.$ Now we differentiate both sides of each equation using implicit differentiation.

$\dd [V]{t} = \frac {\pi }{3}\left (2rh \dd [r]{t} + r^2 \dd [h]{t}\right ) \qquad \text {and}\qquad \dd [r]{t} = \frac {1}{3}\cdot \dd [h]{t}.$ These two equations hold over some time interval. In particular, the equations are true at time when $h=4$ cm. Now we evaluate all the quantities at that time. We plug in $\dd [V]{t} = \answer [given]{10}$ , $r = \answer [given]{4/3}$ , $\dd [r]{t} = \frac {1}{3}\cdot \Bigl [\dd [h]{t}\Bigr ]_{h=4}$ and $h=\answer [given]{4}$ .

$\begin{align*} 10 &= \frac{\pi}{3}\left(2\cdot \frac{4}{3}\cdot 4 \cdot\frac{1}{3}\cdot\Bigl[\dd[h]{t}\Bigr]_{h=4} + \left(\frac{4}{3}\right)^2 \Bigl[\dd[h]{t}\Bigr]_{h=4}\right)\\Now,\hspace{0.025in}\textbf{solve.}\\ 10 &= \frac{\pi}{3}\left(\frac{32}{9}\Bigl[\dd[h]{t}\Bigr]_{h=4} + \frac{16}{9} \Bigl[\dd[h]{t}\Bigr]_{h=4}\right)\\ 10 &= \frac{16\pi}{9}\Bigl[\dd[h]{t}\Bigr]_{h=4}\\ \frac{90}{16\pi} &= \Bigl[\dd[h]{t}\Bigr]_{h=4}. \end{align*}$ Thus, $\Bigl [\dd [h]{t}\Bigr ]_{h=4}=\answer [given]{\frac {90}{16\pi }}$ cm/s.