Asked by Anon
The slope of the tangent to the curve y=f(x) is given by (1/9)(x^2)(y^2). The curve passes through the point (3,1). Find the value of y when x=3*cubedroot(3).
Answers
Answered by
Damon
9 dy/dx = x^2 y^2
9 dy/y^2 = x^2 dx
-9 / y = (1/3) x^3 + c
when x = 3, y = 1
-9/1 = (1/3)(27) + c
-9 = 9 + c
c = -18
so
-9/y = (1/3) x^3 + c
if x = 3 * 3^(1/3) = 3^(4/3)
then x^3 = 3^4
so
-9/y = (1/3)* 3 * 3^3 - 18
-9/y = 27 - 18 = 9
-1/y = 1
y = -1
9 dy/y^2 = x^2 dx
-9 / y = (1/3) x^3 + c
when x = 3, y = 1
-9/1 = (1/3)(27) + c
-9 = 9 + c
c = -18
so
-9/y = (1/3) x^3 + c
if x = 3 * 3^(1/3) = 3^(4/3)
then x^3 = 3^4
so
-9/y = (1/3)* 3 * 3^3 - 18
-9/y = 27 - 18 = 9
-1/y = 1
y = -1
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.