Two Calculus Problem Solving Questions.

[danizh]

#1: The tangent at a point P on the curve y = x^3 intersects the curve again at a point Q. Show that the slope of the tangent at Q is four times the slope of the tangent at P.

So far, I found the derivative of y = x^3, which is y' = 3x^2. Now I set these two equal to each other:

X^3 = 3x^2
x^3 - 3x^2 = 0
x^2 (x-3) = 0
so x = 0 or x = 3.

I don't know where to go from here, or if I did this right at all.

#2: Show that the x- and y- intercepts for any tangent to the curve y = 16 - 8sqrt(x) + x have a sum of 16.

First I expanded the equation:

y = 16 - 8sqrt(x) + x
= 16 - 8^(1/2) + x

Then I determined the derivative:

y' = -4x^(-1/2) + 1
= -4/sqrt(x) + 1

Once again, I have no idea what do from here. :(
Any help would be greatly appreciated.

 

[Fermat]

For #1, assume that it's true. That slope at Q is 4 times slope at P.

P(x_p, y_p)
Q(x_p, y_p)

m_p = 3x²_p
m_q = 3x²_q

and

m_q = 4*m_p

hence

x²_q = 4x²_p

Now write the slope of the line PQ in terms of the coords of P and Q. If you can show that this slope is the same as the slope of the tangent at P, then you will have proved your point. Q.E.D.

 

[Fermat]

For #2,

Let P(x_p, y_p) be a point on the curve.

Get the slope of the tangent at P.

Get the eqn of the tangent at P, in terms of the coords at P.

To get the y-intercept, y_i, set x = 0, in the eqn of the tangent, and solve for y_i.

To get the x-intercept, x_i, set y = 0, in the eqn of the tangent, and solve for x_i.

Now add x_i and y_i and manipulate the resulting expression until it becomes 16.

 

[VietDao29]

Now I set these two equal to each other:
X^3 = 3x^2
x^3 - 3x^2 = 0
x^2 (x-3) = 0
so x = 0 or x = 3.
I don't know where to go from here, or if I did this right at all.

Why are you setting the two equal each other?
That's not correct.
The tangent line at P will be:
g(x) = f'(x_P)(x - x_P) + y_P
Now you have f(x) = x³
To find point Q, x_Q is the solution to the equation:
x_Q³ = f'(x_P)(x_Q - x_P) + y_P = (3x_P²) (x_Q - x_P) + x_P³ = 3x_P²x_Q - 3x_P³ + x_P³ = 3x_P²x_Q - 2x_P³.
<=> x_Q³ - 3x_P²x_Q + 2x_P³ = 0
Now you have to relate x_Q and x_P.
Can you go from here?
Note that the slope of the tangent at P is 3x_P², and the slope of the tangent at Q is 3x_Q².