Related Rates:
Related Rates typically involve problems that can have multiple variables that are
changing in relation to time. Because both of these variables are both related to time,
these problems are called Related Rates problems. Because time is defined implicitly we
will have to use implicit differentiation to find the derivatives of these problems.
Example 1. Suppose that both x and y are both functions of time, t, and that x and y are
related by the equation xy 2  y  x 2  17 .
Suppose further that when x = 2 and y = 3, then dx/dt = 13. Find dy/dt at that moment.
Because x and y are functions of t we will use implicit differentiation and whenever we
need to find dx or dy, they will be replaced by dx/dt or dy/dt
xy 2  y  x 2  17
dy
dx dy
dx
x(2 y )  y 2 
 2x
dt
dt dt
dt
We know that when x = 2 and y = 3 that dx/dt = 13. Just plug in these values and solve
for dy/dt
xy 2  y  x 2  17
dy
dy
2(6 )  32 13 
 4 13
dt
dt
dy
dy
12  117 
 52
dt
dy
dy
13  65
dt
dy
 5
dt
This tells us that y is decreasing by 5 units at that point in time
For the related rates problems it is helpful to:
1) Identify all given quantities and recognize that any values that represent a change
are your dx/dt, dy/dt etc
2) Write an equation relating the variables of the problem
3) Use implicit differentiation to find the derivative with respect to time
4) Solve for the derivative of the unknown rate and substitute in the given values
Example 2
A small rock is dropped into a lake. Circular ripples spread out over the surface of the
water, with the radius of each circle increasing at the rate of 1.5 ft/sec find the rate of
change of the area inside the circle formed by a ripple at the instant the radius is 4 feet.
Given: Area of a circle is A   r 2 ; dr/dt = 1.5; r = 4
Take the derivative of the Area:
A   r2
dA
dr
 2 r
dt
dt
Now substitute in our known values
dA
 2 (4)(1.5)
dt
dA
 12
dt
dA
 37.7
dt
When the radius is 4, the area is increasing by 37.7 ft2/sec
Example 3
A 50 foot ladder is placed against a large building. The base of the ladder is resting on an
oil spill, and it slips at a rate if 3 feet per minute. Find the rate of change of the height of
the top of the ladder above the ground at the instant when the base of the ladder is 30 feet
from the base of the building:
For this problem I would draw a picture and identify all givens. If we sketch the situation
we would have a right triangle where the ladder is the hypotenuse (50’) and the base (30’)
and an unknown height. We can designate the base as x, the unknown height as y, the rate
the base is moving (3 feet per minute) as dx/dt and the height from the ground as dy/dt.
dx
 3; hypotenuse  50
dt
x 2  y 2  50 2
We have:
Remember that we substitute after we derive
dx
dy
Substitute our known values back into the equation
2x  2y
0
dt
dt
dy

2(30)(3)  2y
0
dt
We need y, so we can use Pythagorean theorem 302  y 2  502 , y  40

dy
2(30)(3)  2(40)  0
dt

dy
180  80  0
dt
dy 180


 2.25
dt
80
Given: x  30;
When the base has slid 30’, the top of the ladder is moving downward at a rate of 2.25
feet per minute

Example 4
A cone shaped icicle is dripping from the roof. The radius of the icicle is decreasing at a
rate of 0.2 cm/hr, while the length is increasing at a rate of 0.8 cm/hr. If the icicle is
currently 4 cm in radius and 20 cm long, is the volume of the icicle increasing or
decreasing, and at what rate?
1
Vcone   r 2 h (we will let the length of cone = h)
3
Given: dr/dt = -0.2; dh/dt = 0.8; r = 4; h = 20
Take derivative of the volume:
dV 1  2 dh
dr 
  r
 2rh  Substitute in the given values
dt 3  dt
dt 
dV 1
   42 (0.8)  2(4)(20)( 0.2) 
dt 3
dV 1
  (19.2)  20
dt 3
The volume is decreasing at a rate of 20 cm3 per hour
Example 5
Blood flows faster the closer it is to the center of a blood vessel. According to
Poiseuille’s laws, the velocity V of blood is given by
V  k  R2  r 2  ,
Where R is the radius of the blood vessel, and k is a constant, in this case 375. Suppose a
skiers blood vessel has a radius R = 0.08mm and that cold weather is causing the vessel
to contract at a rate of dR/dt = -0.01 mm/min. How fast is the velocity of the blood
changing? (assume that r2 is also a constant)
V  375( R 2  r 2 )
dV
dt
dV
dt
dV
dt
dV
dt
dR


 375  2 R
 0
dt


dR
 750 R
dt
 750(0.08)(0.01)
 0.06
The velocity of the blood is decreasing at a rate of 0.06 mm/min