for the certain curve, (dˆ2*y) / (dxˆ2) = 18x-8

If the slope of the line tangent to the curve is 9 at the point (1, -1), find the equation of the curve.

Thanks a lot!

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- Mar 1st 2007, 04:27 AMmightymaxCalc problem find the curve.
for the certain curve, (dˆ2*y) / (dxˆ2) = 18x-8

If the slope of the line tangent to the curve is 9 at the point (1, -1), find the equation of the curve.

Thanks a lot! - Mar 1st 2007, 04:36 AMJhevon

We see (dˆ2*y) / (dxˆ2) = 18x-8, this just means the second derivative of some function y is 18x-8, so here goes.

(dˆ2*y) / (dxˆ2) = 18x-8

=> dy/dx = 9x^2 - 8x + C ..................integrated both sides to reverse the derivative

dy/dx gives equation for the slope, which is 9 when x = 1

=> 9 = 9(1)^2 - 8(1) + C

=> C = 8

So dy/dx = 9x^2 + 8x + 8

=> y = 3x^3 - 4x^2 + 8x + D .................again integrated both sides

we have the point (1,-1), so y=-1, when x = 1

=> -1 = 3(1)^3 - 4(1)^2 + 8(1) + D

=> D = -8

So the curve is y = 3x^3 - 4x^2 + 8x - 8 - Mar 1st 2007, 04:43 AMearboth
Hi,

if y = f(x) then

f'(x) = ∫(18x-8) dx = 9x²-8x+C

You know that f'(1) = 9. Therefore

f'(1) = 9 - 8 + C = 9 ===> C = 8 and f'(x) = 9x² - 8x + 8

f(x) = ∫(9x² - 8x + 8)dx = 3x³ - 4x² + 8x + C

You know that f(1) = -1 . Therefore

f(1) = 3 - 4 + 8 + C = -1 ===> C = -8

The term of the function is:

f(x) = 3x³ - 4x² + 8x -8

EB