Related Rate Examples from the Chapter 3, Section 7 Notes
Example 6: A point is moving along a circle x 2 + y 2 = 25 in the first quadrant in such a way that its x
coordinate changes at the rate of 2 cm/sec. How fast is its y coordinate changing as the point passes
through (3, 4)?
We know values for x and y €
at a particular instant, as well as a rate of change for the x-coordinate
(dx/dt = 2 cm/sec). We want to know the rate of change of the y-coordinate (dy/dt). So, it looks
like we have to implicitly differentiate the equation for the circle…
d 2
[ x + y 2 = 25] (1)
dt
So, applying the chain and power rules, we find:
dx
dy
2x
+ 2y
= 0 (2)
dt
dt
And rearranging equation 2 to€solve for what we want (dy/dt), we get:
⎛ dx ⎞
−x⎜ ⎟
dy
⎝ dt ⎠
€
=
(3)
dt
y
So, we can plug in the values (with units!) we know and solve for the rate we want (dy/dt).
dy
3
= − cm/sec.
dt
2
€
Example 7: The dimensions of a rectangle are continuously changing. The width increases at the rate of 3
in/sec while the length decreases at the€rate of 2 in/sec. At one instant the rectangle is a 20-inch square.
How fast is its area changing 3 seconds later? Is the area increasing or decreasing?
In the problem, we’re given a rate of change for the width (dw/dt = +3 in/sec), a rate of change for
the length (dl/dt = -2 in/sec), and that the rectangle is a square (side-length of 20 inches) at one
instant. They are positive/negative rates because we know they are increasing/decreasing.
We’re asked to find the rate of change of its area (dA/dt) exactly three seconds after that instant,
and whether that is increasing or decreasing.
We know that all rectangles have the same formula for area.
A = lw (1)
And we can apply the derivative machine (with respect to time) to equation 1.
d
[ A = lw] (2)
dt
€
Applying the product rule is important, since both l and w are dependent upon time, we get:
dA
dl
dw
(3)
=w +l
dt
dt
dt
€
This equation is true for all rectangles!
For our particular situation, we’re interested in what
happens exactly 3 seconds after the rectangle is a 20-inch square. Therefore, we know that the
length at that point will be 14 inches (since length decreases at a rate of 2 in/sec for 3 seconds);
€ width at that point will be 29 inches (since width increases at 3
and likewise, we know that the
in/sec for 3 seconds).
So, going back to equation 3 and substituting in values, we find:
dA
2
in
in
= (14in )( 3 sec
+ (29in )( −2 sec
= −16 insec (4)
)
)
dt
This gives us the rate of change of the rectangle’s area as negative 16 square inches per second.
The negative means the area is decreasing at that instant.
€
Example 10: A kite is flying 200 ft above the ground, moving in a strictly horizontal direction at a rate of
10 ft/sec. How fast is the angle between the string and the horizontal changing when there is 300 feet of
string out?
I am going to go through a long-winded explanation process on this one, spelling out each step of
the way and my thought process behind it, because I want it to make sense on a step-by-step basis.
In order to get the most out of this, take your time, read this over, and work out the problem along
with me as we go.
We know the height of the kite (vertical distance = 200 ft), and that the kite moves horizontally
(i.e., it’s vertical distance does not change, so dy/dt = 0) at a rate (dx/dt) of 10 ft/sec. We can
assume that the kite-flier is standing still, letting out line at a constant rate. We’re asked to find a
rate of change of an angle (dθ/dt), but a diagram would help a great deal…
It’s no Picasso, but hey… it helps show what we’re looking for.
We want to find dθ/dt so we need an equation that involves angle, hypotenuse, and vertical leg of
a right triangle. Specifically, at the instant of interest, that means we know:
200 ft
sin θ =
(1)
300 ft
Which tells us that the angle at that instant is ~0.7297 radians. This is not the rate of change of the
angle, just the actual angle itself. More generally, we can redraw our diagram more analytically,
€
and rewrite equation 1 as:
y
(2)
h
We can differentiate this equation with respect to time, as follows.
d ⎡
y ⎤
sin θ = ⎥ (3)
⎢
dt ⎣
h ⎦
€
sin θ =
€
⎛ dy ⎞ ⎛ dh ⎞
h⎜ ⎟ − y⎜ ⎟
⎛ dθ ⎞
⎝ ⎠ ⎝ ⎠
(cosθ )⎜ ⎟ = dt 2 dt (4)
⎝ dt ⎠
h
Normally, we’d be good to go at this point – just differentiate and solve for dθ/dt. But, we have a
problem – we know what h is at the instant in question (300 ft of line), but we don’t know what
dh/dt is (it wasn’t a given in the problem). Instead, we have to make use of dx/dt in order to
substitute. So, starting€with the Pythagorean Theorem:
h 2 = x 2 + y 2 (5)
or x = h 2 − y 2 (6)
We can use this equation to figure out the horizontal position of the kite at the moment of interest
(when y = 200 ft and h = 300 ft,€we find that x = 223.6068 ft).
h = x 2 + y 2 (7)
€
We can differentiate equation 7 implicitly with respect to time to get a value for dh/dt:
−1/ 2 ⎛
dh 1 2
dx
dy ⎞
= ( x + y 2 ) ⎜2x
+ 2y ⎟ (8)
⎝ dt
dt 2
dt ⎠
€
Simplifying the result we get:
⎛ dx
dy ⎞
+ y ⎟
⎜ x
dh ⎝ dt
dt ⎠
€
(9)
=
dt
x2 + y2
Now we have all the parts we need in order to solve the problem.
We know how fast the kite travels horizontally (dx/dt = 10 ft/sec), how fast the kit travels
vertically (dy/dt = 0 ft/sec), where the kite is located at the instant of interest (x = 223.6068 ft, and
€ been let out (h = 300 ft), how fast the line being let out at that
y = 200 ft), how much line has
instant (using equation 9, we find that dh/dt = 7.4536 ft/sec), and the angle the kite string makes
with the horizontal (θ = 0.7297 radians).
Therefore, solving equation 4 for dθ/dt (what we’re after), we get:
⎛ dy ⎞ ⎛ dh ⎞
h⎜ ⎟ − y⎜ ⎟
dθ
⎝ dt ⎠ ⎝ dt ⎠
(10)
= (sec θ )
dt
h2
Now we could go ahead and substitute values into this equation and solve for dθ/dt at this point,
but I’m going to continue simplifying to show something important…
Using what we know about the situation (dy/dt = 0) and the definition of secant, we get:
€
⎛ dh ⎞
−y⎜ ⎟
dθ ⎛ h ⎞ ⎝ dt ⎠ −y ⎛ dh ⎞
= ⎜ ⎟
=
⎜ ⎟ (11)
dt ⎝ x ⎠ h 2
hx ⎝ dt ⎠
And now making use of equations 6, 7, and 9, and doing a little algebra, we find:
dθ −y ⎛ dx ⎞
=
⎜ ⎟ (12)
dt h 2 ⎝ dt ⎠
€
I went through all that in order to show you that, even though it seems like it was really
complicated, the solution to the problem only requires the known quantities that we started
with. Everything else was just a matter of substitution and simplification.
€ go, is that the angle appears to be decreasing (dθ /dt is
The punch line, whichever way you
negative) at a rate of 0.02222 radians per second. Recall that “radians” are unit-less, so when
we plug in values into equation 12, we end up with units of “per second” – which makes sense if
we’re measuring an angle in radians.
It is also important to note that if you wait until the end (equation 12) to do your calculations your
answer will be more accurate, since there will be fewer approximations (roundings) along the way.
Therefore, while you could plug values into equation 10 to find the answer, the more accurate
answer is the one obtained by using equation 12.
Example 11: A conical tank (with vertex down) is 18 feet across at the top and 18 feet deep. If the water
is flowing into the tank at a rate of 12 cubic feet per minute, find the rate of change of the depth of the
water when the water is 10 feet deep.
I won’t be as wordy or long-winded here, but we’ll follow the same
general pattern for solving this one.
Known quantities: Tank cone height = 18 ft; Tank cone radius = 9 ft;
Water cone height = 8 ft; Rate of volume increase = 12 ft3/min.
Looking to find: rate of change of the height of the water cone (dh/dt)
at a specific moment when water height is 8 ft.
Equations to use: The volume of a cone. Also, we have two cones (the
tank and the water at the instant in question) that are proportional.
We start with the equation for the volume of a right circular cone (see book’s back inside cover).
V = 13 πr 2 h (1)
Equation 1 can be implicitly differentiated with respect to time (recall only r and h are variables).
dV π ⎡
dr
dh ⎤
= ⎢2rh + r 2 ⎥ (2)
dt
3 ⎣
dt
dt ⎦
€
So, in order to find dh/dt, we need to know the water cone’s radius and dr/dt at the instant the
water cone’s height is 8 feet – we need another equation to use. We can look at a right-triangle
bisection of each cone (see picture above) and set up a proportion since they are similar triangles.
€
radiustan k radiuswater
=
(3)
height tan k height water
From this, we find that the radius of the water cone is 4 feet.
We can also use equation 3 to find dr/dt by using implicit differentiation. Recall that the tank
radius and height are not variables, therefore they do not change over time. So, when we
€ respect to time, we treat the entire left side as a constant.
differentiate equation 3 with
dr
dh
h−r
0 = dt 2 dt (5)
h
Simplifying and solving for dr/dt, we get:
dr r ⎛ dh ⎞
= ⎜ ⎟ (6)
dt h ⎝ dt ⎠
€
So, we now have enough information to solve the problem. Using equations 2 and 6, as well as
the values given in the problem, we can solve for dh/dt.
dh ⎛ dV ⎞⎛ 1 ⎞ 3
ft/sec
= ⎜ ⎟⎜
⎟ =
€
dt ⎝ dt ⎠⎝ πr 2 ⎠ 4 π
The rate is positive, which means that the height of the water (as we would expect) is increasing at
that instant.
€