Section 4.4– Number 5
A wave of wavelength λ traveling in deep water has speed, v, given by
v= k
λ
c
+
c
λ
where c and k are positive constants. As λ varies, does such a wave have a
maximum or minimum velocity? If so, what is it? Explain.
Solution:
Before we find v', let's rewrite v to enable us to find v' a little easier.
1
2
1
v = k  λ + c λ −1 
c

Now using the General Power Rule and the Chain Rule, we have
1 1
−1 
v' = k  
 λ + c λ 
 2  c

−
1
2
1 c 
k − 2 
1
c λ 
−2 
 c − cλ  =
1


 λ c 2
2 + 
c λ
Setting v' equal to 0 to find our critical values
1 c 
k − 2 
c λ  = 0
1
 λ c 2
2 + 
c λ
1 c 
k − 2  = 0
c λ 
1 c
−
=0
c λ2
λ2 − c2 = 0
2
λ2 = c
Multiply by the LCD.
Divide by k.
Multiply by the LCD
λ = +c
But when λ = −c, we have
v= k
−c c
+
= k −1 − 1 = k −2
c −c
OOPS!
So λ = c. But is it a maximum or a minimum? We will use the 1st Derivative test
because, quite frankly, v'' is not going to be easy.
1
2
Since λ = c is our critical value, we will examine λ = c and λ = 2c to determine
where v' is positive (v increasing) and negative (v decreasing).


1 c 
k − 2 
c c 


1 
4  =
v c = 
1
2 
c
2

c
2 2 + 
 c c 

2
1
c 

k −
 c ( 2c )2 

v(2c) = 
1 4
 3
k  −  k − 
c c  =  c
1
1
1
2
 5 2
2 + 2
2 
2

2
< 0 and so v is decreasing.
c 
1
 4 1
3
k − 2  k
k 
− 
4c c 
c 4c 
c
= 
= 
=  1 > 0 and so v is increasing.
1
1
1
1 2
 2c c  2

 5 2
 5 2
2
+
2 2 + 
2 
2 

2
 c 2c 

2
2
Therefore, at λ = c, v is a minimum.