let me know if you want the maximum vertical distance between them.
If you want the distance such that they have a common normal at that place, the solution appears to be a nightmare.
what is the maximum distance between f and g when f(x)=1/2x^2 and g(x)=1/16x^4-1/2x^2 on closed interval [0,4]
3 answers
Reiny, i need the maximum vertical distance between the two functions of the graph.
ahh, not so hard
Did you make a rough sketch?
The two curves intersect at x=0 and x=4, which just happens to be the domain to consider.
vertical height = x^2/2 - (x^4/16 - x^2/2)
= x^2 - x^4/16
d(height)/dx = 2x - 4x^3/16 = 2x - x^3/4
= 0 for max/min of height
x(2 - x^2/4) = 0
x = 0 , which would give a min height
or
2- x^2/4 = 0
x^2 = 8
x = √8 or 2√2
max distance = x^2 - x^4/16
= 8 - 64/16 = 8-4 = 4 units
check my arithmetic
Did you make a rough sketch?
The two curves intersect at x=0 and x=4, which just happens to be the domain to consider.
vertical height = x^2/2 - (x^4/16 - x^2/2)
= x^2 - x^4/16
d(height)/dx = 2x - 4x^3/16 = 2x - x^3/4
= 0 for max/min of height
x(2 - x^2/4) = 0
x = 0 , which would give a min height
or
2- x^2/4 = 0
x^2 = 8
x = √8 or 2√2
max distance = x^2 - x^4/16
= 8 - 64/16 = 8-4 = 4 units
check my arithmetic
1 answer