Calculus - curves and tangent lines. (1 Viewer)

[Forum Friend]

Help please. It's a question which I have difficulty understanding.

"Let V be a point on the curve y=x^3 and suppose the tangent line at V intersects the curve again at W. Prove that the gradient of y=x^3 at W is four times that at V."

Please help explain this and the steps I need to take in order to find the answer, to me.
Thank you.

 

[joey_prince42]

Let V(a,a^3). The gradient at V is 3a^2 so using the point gradient form of a line, the equation of the tangent is

y-a^3=3a^2(x-a) which simplifies to y=3a^2x-2a^3

Now if the the tangent intersects the curve y=x^3, the equation becomes , upon simplification, x^3-(3a^2)x+2a^3=0 .

Now keep in mind that we already have a double root on our hands (since the tangent line has already intersected a point for us). So we only need to find the other real root. There are 4 methods we could use - we could use:

i) polynomial division using the factor x-a (or (x-a)^2) as a divisor (long and tedious)
ii) Differentiate using the fact there is a multiplicity of 2 from the double root (4U method - very easy method)
iii) Use the sum/product of roots (also easy)
iv) Guess

From any of those methods, you should get a real root of x=-2a and upon substituting into the derivative, gives you 3(-2a)^2 = 12a^2 = 4(3a^2) = 4*Gradient at V

I hope that answers your question.

Joey