# Find the curve whose slope at the point (x, y) is 3x^2 if the curve has to pass through the point (1,-1)?

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We know that the slope of a curve y = f(x) at the point (x, y) is given as dy/dx. So here, first we find the general solution of the differential equation by integrating both sides with respect to x:

dy/dx= 3x^2

=> y = Int [3x^2 dx]

= x^3 +C

Now to identify the particular solution whose graph passes through the point (1, -1), we find the value of C for which y= -1 when x = 1

y = x^3 +C

=> -1 = 1^3 +C

=> C = -1-1

=> C= -2

**Therefore the solution we are looking for is y = x^3 – 2.**

The slope of the curve at (x , y) =3x^2.

Therefore the curve must be the premtive of dy/dx = 3x^2.

So the curve is y = Integral (3x^2)dx

y = (3x^3)/4+ c, Or y = x^3+c ,where c is a constant. Since this curve passes through (1 , -1), the coordinates (1 , -1) should satisfy y = x^3+c.

-1 = 1^3+c.

c = -1-1 = -2

Therfore the required curve is y = x^3 -2.