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Transcript
Ch. 5 – Applications of
Derivatives
5.6 – Related Rates
• Ex: Find the derivative of the volume of a sphere with respect to
its radius.
– Differentiate each variable with respect to r!
4
V   r3
3
dV
2 dr
 4 r
dr
dr
dV
 4 r 2
dr
• Ex: Find the derivative of the volume of a sphere with respect to
time.
– Now differentiate each variable with respect to t!
4
V   r3
3
dV
2 dr
 4 r
dt
dt
– This equation relates two rates, dV/dt and dr/dt, so this problem is a related
rates problem!
– Related Rates problems are useful when observing the change in variables
over time!
• Ex: Find an equation that relates the volume of a cylinder over time
with its radius and height over time.
– Use Product Rule!
V   r 2h
dV
dr 

2 dh
 ( r )   2 r  h
dt
dt 
dt 
dV
dr
2 dh
r
 2 rh
dt
dt
dt
• Ex: A spherical kickball is being inflated. When the radius is 3 in.,
the volume is changing at a rate of 2π in/s. What is the rate at
which the radius is growing when the radius is 3 in?
– Always label what you know and what you need to find...
– ...then set up an equation that relates all of your rates. You’ll probably need
to differentiate an existing equation with respect to t.
4 3
V  r
3
Know
@ r  3 in,
dV
in
 2
dt
s
dV
2 dr
 4 r
dt
dt
Find
dr
@r 3
dt
– Plug in the known values ONLY AFTER DIFFERENTIATING. Otherwise, you
will be differentiating constants, not variables.
dr
 2   4 (3)
dt
2
dr 1 in

dt 18 s
• Ex: A man is 18 m away from a child. The child releases a balloon
that rises at a rate of 4 m/s. How quickly is the angle of elevation
from the man to the balloon increasing 6 s after the child releases
the balloon?
B
– Draw and label a picture to help you!
– Find an equation that relates your rates...
Know
@ t  6 s,
dh
m
4
dt
s
h
tan  
18
θ
M
d 1 dh
(sec  )

dt 18 dt
2
– At t = 6 , the height of the balloon will be...
– ...24 m, and using tanθ = 24/18, θ will be...
– ...≈.927 radians! Now plug n’ chug!
d 1
sec (.927)
 (4)
dt 18
2
h
d
radians
 .08
dt
s
18 m
C
Find
d
@t 6
dt