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Mathematical Expression Editor
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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 . First, the
cook stretches the dough to the shape of a cylinder of radius 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 cm/min, how fast is its thickness changing when the radius is
cm?
First, we introduce the variables , , and , denoting the volume, the radius,
and the thickness of the pizza, in that order. We identify the given rate cm/min
and the unknown rate , when .
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,
and note that , and write which immediately simplifies to Since both and are
functions of time, we now differentiate both sides of the equation using implicit
differentiation,
Now we’ll evaluate all the quantities at the instant when . But, first, we have to
compute at that moment. Since it follows that , when . Now, we are ready to
solve.
Therefore, when ,
Hence, the thickness of the dough is changing at a rate of cm/min when
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 , , and , denoting the volume, the radius, and the
surface area of the snowball, in that order . Then, we identify the given rate and
the unknown rate , 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, 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 . Since is a function of , we differentiate both sides of equation .
So On the other hand, Therefore, Now, we solve for and obtain that
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 . Cyclist
is riding north along the first road, and cyclist is riding east along the
second road. At a particular time, cyclist is kilometers to the north of and
traveling at km/h, while cyclist is kilometers to the east of and traveling at
km/h. How fast is the distance between the two cyclists changing at that
time?
First, we introduce the variables , , and , denoting the distance of cyclist
from the point , the distance of cyclist from the point , and the distance
between the two cyclists, in that order. We identify the given rates km/h,
km/h, when and the unknown rate , when . Now, we draw a picture.
We find equations relating the variables , , and . By the Pythagorean Theorem,
Since , , and are functions of time, we differentiate both sides of the equation with
respect to . This equation holds on some time interval, and , in particular, at the
instant when and . Therefore, at that instant,
In order to evaluate , the distance between the cyclists, at the instant when and , we
draw a picture “taken” at that instant.
By the the Pythagorean Theorem and the picture above, when , , it follows that
Now we can evaluate all the quantities in equation and solve We find that km/hr
at the moment when , .
A plane is flying at an altitude of miles directly away from you at 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 miles from you?
First, we introduce variables , the distance the plane has traveled after it flew
right above you, and , the distance between you and the plane. The given
rate is mph and the unknown rate is , when . Next, we draw a picture.
Next we find equations relating the variables and . By the Pythagorean Theorem
we know that Since and are functions of time, we now differentiate both sides of
the equation At the instant when mi, we get that This implies that has to be
evaluated at time when . In order to do that, let’s draw the picture at the given
instant, when mi.
So, . Putting together all the information we get Finally, we solve: mph, at the
moment when miles.
Angular rates
A plane is flying at an altitude of miles directly away from you at mph . Let 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 changing at the
moment when the plane is flying over a point on the ground miles from
you?
First, we introduce a variable , the distance the plane has traveled after it flew
right above you. So, the given rate is mph and the unknown (related) rate is , at
the instant when miles. This rate should be measured in rad/s. Therefore,
we have to convert the units of the given rate, mph, into mi/s: mph mi/s.
Next, we draw a picture.
Now we find an equation relating variables and . From the picture we can see that
Since and are both functions of time, we now differentiate both sides of the
equation. We write The above equation holds over some interval of time. In
particular, it is true when . Therefore, when , This implies that we have to compute
the unknown value, , when . To this end, we draw and label the picture at the time
when mi.
So, at the moment when miles, Now, by substituting all the known values into the
equation above, we get We solve for , when , and get that
Similar triangles
It is night. Someone who is feet tall is walking away from a street light at a rate of
feet per second. The street light is 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 feet from the street light?
First, we introduce two variables, , the distance from the person to the lamp, and
, the length of the shadow. The given rate is ft/s and the unknown (related) rate is
at the moment when feet. Next, we draw a picture.
Now we find equations that relate the variables and . We use the fact that we have
similar triangles to write:
Since and are both functions of time, we differentiate both sides of the equation
above. We use implicit differentiation and write At this point we evaluate all
the quantities at time when ft Now we solve for and get that , when
ft.
Therefore, the shadow is growing at a rate of ft/s when the person is ft from the
lamp.
Water is poured into a conical container at the rate of 10 cm/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 be the volume of the water in
the container, let be the radius of the circular surface of the water, and
let be the depth of the water in the container. The given rate is , and
the unknown rate is , at the moment when 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 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 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 in terms of
.
QUESTION: Why is it more convenient to express in terms of than in terms of
?
Therefore, and Now, we differentiate with respect to . This equation holds over
some time interval. In particular, the equation is true at time when cm. Now
we evaluate all the quantities at that time. So, at the instant when the
water in the container is cm deep, the water level is rising at the rate of
cm/s.