Length of the curve calculus?
2019-03-20 4:13 pm
Is it the final option? Or am I wrong, if i am, can you explain please! I took a picture of the question
回答 (2)
2019-03-20 4:56 pm
x = (1/3) * (y^2 + 2)^(3/2)
x' = (1/3) * (3/2) * 2y * (y^2 + 2)^(1/2) = y * (y^2 + 2)^(1/2)
sqrt(1 + (x')^2) * dy =>
sqrt(1 + y^2 * (y^2 + 2)) * dy =>
sqrt(y^4 + 2y^2 + 1) * dy =>
sqrt((y^2 + 1)^2) * dy =>
(y^2 + 1) * dy
Integrate from y = 0 to y = 1
(1/3) * y^3 + y + C =>
(1/3) * (1^3 - 0^3) + (1 - 0) + C - C =>
(1/3) * 1 + 1 =>
4/3
x' = (1/3) * (3/2) * 2y * (y^2 + 2)^(1/2) = y * (y^2 + 2)^(1/2)
sqrt(1 + (x')^2) * dy =>
sqrt(1 + y^2 * (y^2 + 2)) * dy =>
sqrt(y^4 + 2y^2 + 1) * dy =>
sqrt((y^2 + 1)^2) * dy =>
(y^2 + 1) * dy
Integrate from y = 0 to y = 1
(1/3) * y^3 + y + C =>
(1/3) * (1^3 - 0^3) + (1 - 0) + C - C =>
(1/3) * 1 + 1 =>
4/3
收錄日期: 2021-05-01 22:31:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20190320081324AAsXYBC