Homework Help

Given dy/dx=1/square root(36-x^2). What is y ?

user profile pic

guteskind | Student | eNotes Newbie

Posted May 19, 2011 at 4:17 PM via web

dislike 0 like

Given dy/dx=1/square root(36-x^2). What is y ?

Tagged with calculus, math

1 Answer | Add Yours

user profile pic

giorgiana1976 | College Teacher | Valedictorian

Posted May 19, 2011 at 4:23 PM (Answer #1)

dislike 0 like

To determine the primitive function y, we'll have to compute the indefinite integral of the function.

If dy/dx = 1/sqrt(36-x^2) => dy = dx/sqrt[(6)^2 - x^2]

We'll integrate both sides:

Int dy = Int dx/sqrt[(6)^2 - x^2]

We'll recognize the formula:

Int dx/sqrt(a^2 - x^2) = arcsin (x/a) + C

Let a = 6 => Int dx/sqrt[(6)^2 - x^2] = arcsin (x/6) + C

The required function y, when dy/dx = 1/sqrt(36-x^2), is: y = arcsin (x/6) + C.

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes