y^2 = 64(x + 4)^3, 0 ≤ x ≤ 2, y > 0. Created by Rose. science-mathematics-en - mathematics-en

Find the exact length of the curve. y^2 = 64(x + 4)^3, 0 ≤ x ≤ 2, y

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May 2021 1 53 Report

Find the exact length of the curve. y^2 = 64(x + 4)^3, 0 ≤ x ≤ 2, y > 0?

y^2 = 64(x + 4)^3, 0 ≤ x ≤ 2, y > 0

Science & Mathematics / Mathematics

Answers & Comments

queenfan2012

I tried my best, and I got y=x+8. I don't think it's right though. I applied the power of 3 to the parenthesis, then the 4, but that might not have been right. Idk if I should have applied the 64 first.

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