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?
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
To find the x-value of the point of location of the particle after one hour, we need to determine the change in x after one hour.
First, we need to convert one hour to minutes. Since there are 60 minutes in an hour, one hour is equal to 60 minutes.
Next, we need to determine the distance traveled by the particle in 60 minutes. We know that the particle moves at a constant rate of 3 units per minute, so the distance traveled is 3 units/minute × 60 minutes = 180 units.
Now, let's find the initial x-value by substituting x = 1 into the given function:
y = (8/3)x^(3/2)
y = (8/3)(1)^(3/2)
y = (8/3)(1)
y = 8/3
So, the initial point on the graph is (1, 8/3).
To find the x-value of the point after traveling 180 units, we need to find the x-value that corresponds to a y-value of 8/3 + 180.
Solving the function equation for y = 8/3 + 180:
8/3 + 180 = (8/3)x^(3/2)
Multiplying both sides of the equation by 3/8 to isolate x^(3/2):
(8/3 + 180)(3/8) = x^(3/2)
Simplifying:
(8 + 540)/8 = x^(3/2)
(548/8) = x^(3/2)
Now, we need to find the x-value by taking the square of both sides:
[(548/8)^(2/3)]^(3/2) = (x^(3/2))^(2/3)
Simplifying:
(548/8)^(2/3) = x
Finally, we can calculate the numerical value of x using a calculator and rounding the result to the nearest hundredth. The value of x should be approximately 4.36.