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!
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
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