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, 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 our next problem, we will stretch and flatten a cylinder-shaped pizza dough.

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 $h$ , denoting the volume, the radius, and the thickness of the pizza, in that order. We identify the given rate $\frac {dr}{dt}=15$ cm/min and the unknown rate $\frac {dh}{dt}$ , when $r=20$ .

Next, we draw a picture. Note that our pizza dough is shaped as a cylinder.

Next we need to find equations. Recall the formula for the volume of the cylinder, $V=\pi \cdot r^2\cdot h$ and note that $V=400\pi \, \text {cm}^3$ , and write $400\pi = \pi \cdot r^2 \cdot h,$ which immediately simplifies to $400 = r^2 \cdot h.$ Since both $r$ and $h$ are functions of time, we now differentiate both sides of the equation using implicit differentiation,

$0 = 2\cdot r \cdot \frac {dr}{dt} \cdot h+ r^2 \cdot \frac {dh}{dt}.$ Now we’ll evaluate all the quantities at the instant 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.

$0 = 2\cdot 20 \cdot 15 \cdot 1 + 20^2 \cdot \frac {dh}{dt}$ $\frac {-2\cdot 20 \cdot 15 \cdot 1}{20^2} = \frac {dh}{dt}$ Therefore, when $r=20$ ,

$\frac {dh}{dt} = -1.5.$ Hence, the thickness of the dough is changing at a rate of $\answer [given]{-1.5}$ cm/min when $r=20$ cm.

In our next example we consider a melting snowball.

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 $A$ , denoting the volume, the radius, and the surface area of the snowball, in that order . 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.

In order to find equations that relate all relevant variables, we recall formulas for the volume and the surface area of the sphere, $(1) \hspace {0.1in} V =\frac {4}{3} \cdot \pi \cdot r^3 \qquad \text {and}\qquad (2) \hspace {0.1in} A = 4 \pi \cdot r^2.$ 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 $\frac {dV}{dt}$ . Since $r$ is a function of $t$ , we differentiate both sides of equation $(1)$ . So $\frac {dV}{dt} = 4\cdot \pi \cdot r^2 \cdot \frac {dr}{dt}.$ On the other hand, $\frac {dV}{dt} = k\cdot A.$ Therefore, $4\cdot \pi \cdot r^2 \cdot \frac {dr}{dt} = k\underbrace {\cdot 4\cdot \pi \cdot r^2}_{A} .$ Now, 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/h, while cyclist $B$ is $4$ kilometers to the east of $P$ and traveling at $15$ km/h. How fast is the distance between the two cyclists changing at that time?

First, we introduce the variables $a$ , $b$ , and $c$ , denoting the distance of cyclist $A$ from the point $P$ , the distance of cyclist $B$ from the point $P$ , and the distance between the two cyclists, in that order. We identify the given rates $\frac {da}{dt}=20$ km/h, $\frac {db}{dt}=15$ km/h, when $a=3, b=4$ and the unknown rate $\frac {dc}{dt}$ , when $a=3, b=4$ . Now, we draw a picture.

We find equations relating the variables $a$ , $b$ , and $c$ . By the Pythagorean Theorem, $(1)\hspace {0.2in}c^2=a^2+b^2.$ Since $a$ , $b$ , and $c$ are functions of time, we differentiate both sides of the equation with respect to $t$ . $(2)\hspace {0.2in}2\cdot c\cdot \frac {dc}{dt}=2\cdot a\cdot \frac {da}{dt}+2\cdot b\cdot \frac {db}{dt}.$ This equation holds on some time interval, and , in particular, at the instant when $a=3$ and $b=4$ . Therefore, at that instant,

$(3)\hspace {0.2in}2\cdot c \cdot \underbrace {\frac {dc}{dt}}_{unknown\hspace {0.015in} rate} =2\cdot 3\cdot \underbrace {20}_ {given\hspace {0.02in} rate}+2\cdot 4 \cdot \underbrace {15}_{given\hspace {0.02in} rate} .$

In order to evaluate $c$ , the distance between the cyclists, at the instant when $a=3$ and $b=4$ , we draw a picture “taken” at that instant.

By the the Pythagorean Theorem and the picture above, when $a=3$ , $b=4$ , it follows that $c= \answer [given]{5} km.$ Now we can evaluate all the quantities in equation $(3)$ and solve $2\cdot 5 \cdot \frac {dc}{dt} = 2 \cdot 3\cdot 20 + 2 \cdot 4 \cdot 15.$ We find that $\frac {dc}{dt} = 24$ km/hr at the moment when $a=3$ , $b=4$ .

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 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 rate is $\frac {ds}{dt}$ , when $p=4$ . 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 $2\cdot p\cdot \frac {dp}{dt} = 2\cdot s \cdot \frac {ds}{dt}.$ At the instant when $p=4$ mi, we get that $2\cdot 4\cdot \underbrace { 500}_{given\hspace {0.02in}rate} = 2\cdot \overbrace {s}^{unknown\hspace {0.02in}value} \cdot \underbrace {\frac {ds}{dt}}_{unknown\hspace {0.02in}rate} .$ 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 given instant, when $p=4$ mi.

So, $s=\answer [given]{5}$ . Putting together all the information we get $2(4)(500)=2(5)\frac {ds}{dt} .$ Finally, we solve: $\frac {ds}{dt}=\answer [given]{400}$ mph, at the moment when $p=4$ miles.

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 $\theta$ changing 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 it flew right above you. So, the given rate is $\frac {dp}{dt}=500$ mph and the unknown (related) rate is $\frac {d\theta }{dt}$ , at the instant when $p=4$ miles. 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 $\sec ^{2}{\theta }\cdot \frac {d\theta }{dt} = -\frac {3}{p^2}\cdot \frac {dp}{dt}.$ The above equation holds over some interval of time. In particular, it is true when $p=4$ . Therefore, when $p=4$ , $\overbrace {\sec ^{2}{\theta }}^{unknown\hspace {0.02in}value}\cdot \underbrace {\frac {d\theta }{dt} }_{unknown\hspace {0.02in}rate} = -\frac {3}{4^2}\cdot \underbrace { \frac {500}{3600}}_{given\hspace {0.02in}rate}.$ This implies that we have to compute the unknown value, $\sec ^{2}{\theta }$ , when $p=4$ . To this end, we draw and label the picture at the time when $p=4$ mi.

So, at the moment when $p=4$ miles, $\sec ^2{\theta }=1+\Bigr (\frac {3}{4}\Bigl )^2=\frac {25}{16}.$ Now, by substituting all the known values into the equation above, we get $\frac {25}{16}\cdot \frac {d\theta }{dt}= -\frac {3}{16}\cdot \frac {500}{3600}.$ We solve for $\frac {d\theta }{dt}$ , when $p=4$ , and get that $\frac {d\theta }{dt} = -\frac {1}{60} rad/s.$

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, $p$ , the distance from the person to the lamp, and $s$ , the length of the shadow. The given rate is $\frac {dp}{dt}=3$ ft/s and the unknown (related) rate is $\frac {ds}{dt}$ at the moment when $p=7$ feet. Next, 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:

$\begin{align*} \frac{s+p}{15} &= \frac{s}{\answer[given]{6}},\\ 6\cdot s + 6 \cdot p &= 15\cdot s,\\ 6\cdot p &=9\cdot s,\\ 2\cdot p &=3 \cdot s. \end{align*}$ Since $p$ and $s$ are both functions of time, we differentiate both sides of the equation above. We use implicit differentiation and write $2\cdot \frac {dp}{dt} =3 \cdot \frac {ds}{dt}$ At this point we evaluate all the quantities at time when $p=7$ ft $2\cdot \cancel {3} = \cancel {3} \cdot \frac {ds}{dt}.$ Now we solve for $\frac {ds}{dt}$ and get that $\frac {ds}{dt}= \answer [given]{2}$ , when $p=7$ ft.

Therefore, the shadow is growing at a rate of $2$ ft/s 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. Let $V$ be the volume of the water in the container, let $r$ be the radius of the circular 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 $\frac {dh}{dt}$ , at the moment when $h=4$ cm. 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 $\frac {r}{h}=\frac {10}{30} \qquad \text {so}\qquad r=\frac {h}{\answer [given]{3} }.$ At this point we could differentiate both sides of each equation, but we can take a simpler approach by combining these two equations and expressing $V$ in terms of $h$ .

QUESTION: Why is it more convenient to express $V$ in terms of $h$ than in terms of $r$ ?

Therefore, $V = \frac {\pi }{3}\overbrace { \Bigl (\frac {h}{3}\Bigr )^2}^{r^2} h,$ and $V = \frac {\pi }{27} \cdot h^3.$ Now, we differentiate with respect to $t$ . $\dd [V]{t} = \frac {\pi }{27} \cdot 3\cdot h^2\cdot \frac {dh}{dt}.$ This equation holds over some time interval. In particular, the equation is true at time when $h=4$ cm. Now we evaluate all the quantities at that time. $10 = \frac {\pi }{27} \cdot 3\cdot 4^2\cdot \frac {dh}{dt}.$ So, at the instant when the water in the container is $4$ cm deep, the water level is rising at the rate of $\frac {45}{8\pi }$ cm/s.