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DIFFERENTIATION HELP PLEASE!!?
The curve C is given by the equation y=mx^3-x^2+8x+2, for a constant m.
a) Find dy/dx
The point P lies on C, and has the x-value 5. The normal to C at P is parallel to the line given by the equation y+4x-3=0.
b) Find the gradient of curve C at P.
Hence or otherwise, find:
c i) the value of m
c ii)the y-value of P.
Thanks help will be much appreciated!!
2 Answers
- 9 years agoFavorite Answer
a) dy/dx = 3mx^2 - 2x +8
b) substitute 5 into the equation for dy/dx which will give you the gradient in terms of m.
ci) use the fact that the normal to C at P has a gradient of -(dy/dx)^-1 at x = 5. since the gradient of the normal is -4, dy/dx must equal 1/4 at x=5. You can work m out from there.
cii) use your value of m from (ci) in order to work out y using your original equation.
I think this is how to do it, good luck!
