calculus homework help

Question

Find the length of the curve y = x^3/2 on the interval from interval from x = 0 to x =8. Check the interview calculation using a TI-83 or TI-84 calculator.

Answer 1

y=x^3/2 = x^1.5 so y' = 1.5 x^.5

Arc length is given as L=∫√[1+(y')²]dx in this case with bounds from x=0 to x=8

L=₀∫⁸√[1+(1.5x^.5)²]dx=₀∫⁸√[1+2.25x]dx=₀∫⁸(1+2.25x)^1/2dx

use u=(1+2.25x) so x=0 gives u= 1+2.25(0)=1

and x=8 gives u=1+2.25(8)= 1+18 =19

L=₁∫¹⁹ u^1/2du = [u^3/2]¹⁹₁ =(19)^3/2- 1^3/2

Answer 2

Hello! For this question you would need to use the Arc Length formula which states L=∫√1+[f'(x)]dx. The f'(x) is just the derivative of the equation y=x^3/2 while the interval (0,8) is just (a,b). Hope this helps!

Answer 3

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