# 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.

Related Calculus and Beyond Homework News on Phys.org

#### Fermat

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

P(xp, yp)
Q(xp, yp)

mp = 3x²p
mq = 3x²q

and

mq = 4*mp

hence

q = 4x²p

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

Homework Helper
For #2,

Let P(xp, yp) 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, yi, set x = 0, in the eqn of the tangent, and solve for yi.

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

Now add xi and yi and manipulate the resulting expression until it becomes 16.

#### VietDao29

Homework Helper
danizh said:
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'(xP)(x - xP) + yP
Now you have f(x) = x3
To find point Q, xQ is the solution to the equation:
xQ3 = f'(xP)(xQ - xP) + yP = (3xP2) (xQ - xP) + xP3 = 3xP2xQ - 3xP3 + xP3 = 3xP2xQ - 2xP3.
<=> xQ3 - 3xP2xQ + 2xP3 = 0
Now you have to relate xQ and xP.
Can you go from here?
Note that the slope of the tangent at P is 3xP2, and the slope of the tangent at Q is 3xQ2.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving