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The distance (s) in inches through which a cam follower moves in 4s is:

s = ∫4,0 (t)sqrt[4+9t^2]dt

elavulate s.

the two numbers after the ∫ are the upper and lower limits of the itergral.

2007-02-25 02:05:04 · 1 answers · asked by babblefish186 3 in Education & Reference Homework Help

1 answers

s = ∫(t)sqrt[4+9t^2]dt

Use u substitution. We want to use 4 + 9t^2, because the derivative of 4 + 9t^2 will let us get rid of the t.
let u = 4 + 9t^2
du = 18t dt
1/18 du = t dt

That gives you:
s = ∫sqrt[u] * 1/18 du
s = 1/18 ∫u^1/2 du
s = 1/18 u^3/2 / 3/2 | 4,0

Now plug the value of u in for u, and solve:
s = 1/12 (4 + 9t^2)^3/2 | 4,0
s = 1/12 (4 + 9*16)^3/2 - 1/12 (4)^3/2
s = 150.04 - 2/3 = 149.37 in. (solution!)

2007-02-26 05:32:42 · answer #1 · answered by ³√carthagebrujah 6 · 0 0

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