RELATED RATES & IMPLICIT
DIFFERENTIATION
Skill #1: Leibniz notation and function notation
f ( x ) means the derivative of a function f ( x ) , whose input variable is x. Suppose the
dy
output variable of this function is y, i.e. y  f ( x) . Then in Leibniz notation,
is the
dx
same as f ( x ) .
dV
means the derivative of a function whose input variable is t, and whose output
dt
variable is V. In function notation, we could invent a function name, say g , and write
dV
V  g (t ) to express the function. Then
would correspond to g (t ) . Or, often,
dt
instead of using “ g ” as the function name, we use “ V ”, and write “ V (t ) ” as the
function, and “ V (t ) ” as the derivative.
You try the following:
1. If R  f ( y ) , write the derivative in both function notation and Leibniz notation.
dT
2. If there is a derivative
, write the original function and the derivative function
dP
in function notation.
Skill #2: Differentiating “Unknown” functions
Example: y   f ( x)  ; Find
3
dy
dx
Solution: The derivative of “something” cubed is three times “something” squared, times
dy
2
 3  f ( x)   f ( x)
the derivative of “something”:
dx
Example: A  3 tan( r ) ; Find
dA
dr
Solution: Actually this isn’t an example of differentiating “unknown” functions, because
dA
tells us that “r” is the input variable, and “A” is the output variable, so this problem
dr
dA
 3sec 2 ( r )
is exactly like y  3 tan( x) . Thus the solution is
dr
Example: A  3 tan( r ) ; Find
dA
.
dt
Solution: Do you see the difference between this problem and the previous one? Here
dA
tells us that we are looking for “the derivative of A with respect to t”, i.e. t is the
dt
input variable. This means we need to re-write the equation so that every variable is a
function of t: A(t )  3 tan(r (t )) . Now we differentiate, with respect to t. Notice that now
there is something “inside” the tan function, so we need to use the chain rule when
differentiating the right side of the equation: A(t )  3sec2 (r (t )) r (t ) . This could be redA
dr
 3sec 2 r 
written using Leibniz notation as:
dt
dt
You try the following:
dy
. Remember to use the product rule.
dx
dQ
2. Q  3m6  4 ; find
.
dm
dQ
3. Q  3m6  4 ; find
.
dt
f ( x)
1. y  x  e  ; find
dA
. Write the answer
dt
using Leibniz notation. (Remember to first re-write all variables as functions of “t”. You
will need to use the product rule to differentiate this problem.)
4. A  s 2  h . Both s and h are variables, not constants. Find
Skill #3 Identifying Derivatives and Derivative Notation and Understanding Derivatives
Terminology
Example: A sphere is expanding. Find the rate of change of Volume with respect to
radius.
Solution: This terminology means that we are viewing Volume as a function of Radius,
and want to find the derivative. Thus “r” is the input variable (horizontal axis), “V” is the
output variable (vertical axis), and the “rate of change of Volume with respect to radius”
dV
is the derivative, symbolized as
. The units would be in m3/m, meaning that if the
dr
dV
radius increases by 1m, then the volume will increase by approximately
cubic
dr
dV
4
4
meters. To actually find
, we know that V   r 3 . This is just like y   x 3 ,
3
3
dr
dV
because
means that “r”is the input variable and “V” is the output variable. So the
dr
dV 4
   3r 2  4 r 2
derivative is just
dr 3
Example: A sphere is expanding. Find the rate of change of Volume with respect to
Time.
Solution: This terminology means that we are viewing Volume as a function of Time,
and want to find the derivative. Thus “t” is the input variable (horizontal axis), “V” is the
output variable (vertical axis) and the “rate of change of Volume with respect to Time” is
dV
the derivative, symbolized as
. The units would be in m3/s, meaning that if time
dt
dV
increases by 1 second, Volume will increase by approximately
cubic meters. We
dt
4
know that V   r 3 , but notice that we want TIME, not radius, to be the input variable.
3
Therefore we should re-write this equation thinking of everything as a function of TIME:
4
3
2
V (t )    r (t )  . Thus V (t )  4  r (t )   r (t ) . Using Leibniz notation,
3
dV
dr
 4 r 2 
dt
dt
Example: A jogger is running at 3 m/s at a certain moment.
Solution: Clearly this is a velocity. We know velocity is the derivative of Position with
respect to Time, that is, velocity is the derivative of a position – time function (time on
the horizontal axis, position on the vertical axis). Note that the UNITS (m/s) are a
giveaway as well, because the unit on the bottom (seconds) is from the input variable
(time) and the unit on the top (meters) is from the output variable (position). So let’s
define a variable “P” to mean the position of the jogger, measured from some reference
dP
 3 m/s. This means that if t increases by 1, then P will
position. Then we have
dt
increase by approximately 3 meters.
You try:
1. A student is spinning at a rate of 4 radians per second. Define variables clearly,
and write this rate as a derivative using Leibniz notation.
2. As you dive, the rate of change of pressure with respect to depth is 980 Pascals
per meter. What does this sentence mean? Define variables clearly, and write
this rate as a derivative using Leibniz notation.
3. A square is growing. Find the rate of change of area with respect to side length.
Define variables, and express this derivative using Leibniz notation. You should
also be able to calculate this derivative, because you know that Area = (sidelength)2
4. A square is growing. Find the rate of change of area with respect to time. This is
only slightly different in wording from the previous problem, but the difference is
significant.
Skill # 4 – Using the chain rule to relate derivatives to each other
Example: An archer pulls back on the bowstring. As the bow bends, the force exerted by
the archer increases. Identify three variables, three functions, and three rates
(derivatives) and use the chain rule to write a relationship between the three rates.
Solution: Time, the bowstring being stretched, and force are three changing variables.
Time (t, seconds) and Force (F, newtons) are standard concepts, but what about the
stretching bowstring? Let’s measure this in the number of centimeters the string is pulled
back (S, cm).
S is a function of time; F is a function of S; F is a function of time.
The three rates are:
dS dF dF
dF dF dS
,
,


. Their relationship is:
dt dS dt
dt dS dt
Thus, for example, if at any given moment we know any two of these rates, we can find
the third rate easily.
You try:
Suppose a spectator at a bowling game is watching the ball roll down the lane. In order
to do this, the spectator’s head must rotate.
a) Identify three variables, three functions (i.e. say what is a function of what) and
three rates (derivatives) in Leibniz notation, and use the chain rule to write a
relationship between the three rates.
b) Suppose at a certain moment the bowling ball is rolling at 2 m/s. At the same
moment, the spectator’s eyes are rotating at 5 degrees per second. Find the value
of the third rate (using correct units) and explain what it means.
Skill #5 – Putting it all together to solve related rates problems
A) Set up any necessary variables.
B) Identify any rates known, or desired to be known, as derivatives using Leibniz
notation.
C) Sometimes you will be able to solve the problem at this point using skill # 4, because
the rates will relate simply using the chain rule. If not, keep going:
D) Identify any known equations linking any of the variables together.
E) Decide what variable you need to differentiate with respect to, i.e. what you need the
variables to be functions of, and re-write the known equation(s) appropriately.
F) Differentiate with respect to the identified variable.
G) Plug in any known rates and known quantities at the end, and solve for the unknown
rate.
Example: At a certain moment a particle is at the point x=3, y=4, and the point is moving
upward at a rate of 3m/s and moving right at a rate of 2 m/s. Find the rate at which the
distance of the point from the origin is changing.
Draw a diagram of the above situation before reading on. Do so now!
A) Define x and y to be the x- and y- coordinates of the particle at any time t. Notice
that x and y are really functions of t. Define D to be the distance between the
point and the origin. So there are 4 variables, D, x, y, t.
B) The rate 3m/s is talking about how fast “y” (the position vertically) is changing
dy
dx
 3 m/s. Likewise,
 2 m/s. The rate asked for is
with respect to time, so
dt
dt
dD
how fast D is changing with respect to time, so it’s
that is needed.
dt
C) These four variables don’t relate simply using the chain rule (skill #4).
D) We know that D 2  x 2  y 2 by Pythagoras’ theorem.
E) We want to differentiate with respect to time, t, because our known and unknown
2
2
2
rates are all derivatives with respect to t. So we write  D(t )    x(t )    y (t ) 
F) Differentiating with respect to t on both sides yields:
2 D(t )  D(t )  2 x(t )  x(t )  2 y (t )  y (t ) , or, in Leibniz notation,
dD
dx
dy
2D 
 2x   2 y 
dt
dt
dt
dD
G) We known all of the rates and variables except for D and
. To find D, use
dt
D 2  x 2  y 2 at the moment when x=3, y=4. This gives D=5. Plugging
everything in gives:
dD
dD 36
2(5)
 2(3)(2)  2(4)(3) . Thus

 3.6 m/s
dt
dt 10