y = 8/(3x^(3/2))
y' = -4/x^(5/2)
so, the arc length involved is
∫[1,h] √(1+16/x^5) dx
Hmmm. I suspect you meant
y = 8/3 x^(3/2)
y' = 4√x
∫[1,h] √(1+16x^2) dx
Using the trig substitution 4x = sinhθ, that integral is
s = x/2 √(1+16x^2) + 1/8 sinh(4x)
we want s(x) - s(1) = 180
s(1) = 5.47279
I don't know of any method except graphical or numeric to find when s(x)=185.47279, but wolframalpha says it's when
x = 1.9879
A particle moves along the graph of the function y= 8/3x^(3/2) at the constant
rate of 3 units per minute. The particle starts at the point where x = 1 and travels in the
direction of increasing x. After one hour, what is the x-value, rounded to the nearest
hundredth, of the point of the location of the particle?
1 answer
1 answer