Beginning Calculus: Prove Slope of Tangent at Q is 4x Slope at P

[Seinfeld4]

I'm a beginning calculus student and this is one of the 'challenge' questions from an older edition of the Stewart's Calculus series of textbooks.

I believe I did the question properly, but the solution in the book is completely different from my own attempt. I'm not good enough at math yet to feel confident that my solution has no logical flaws or unwarranted assumptions. I would very much appreciate it if someone could have a look at my solution and let me know if it's mathematically sound.

Homework Statement

Let P be a point on the curve y = x^3 and suppose the tangent line at P intersects the curve again at Q. Prove that the slope of the tangent at Q is four times the slope of the tangent at P.

The Attempt at a Solution

Let the x-coordinate of P be "a", and let the x-coordinate of Q be "b". Also, let the slope of the tangent at P be "mL", and let the slope of the tangent at Q be "mK".

1. First I considered the tangent at P to be a secant line since it intersects the graph at two points. Using the formula for the slope of the secant, (f(x + h) - f(x))/h, I got:

(f(b + h) - f(b)) / h = mL
3b^2 + 3bh + h^2 = mL

And, since h is the distance from a to b, h = |a - b|. Consider that a > b, and so |a - b| = a - b. For the case in which b > a, the slope of the secant becomes 3a^2 + 3ah + h^2, and h = b - a. The leaves the final solution unaffected. Also, mL = 3a^2, since that is the value of the derivative at a.

Substituting these values into the above equation and simplifying yields:

b^2 + ab - 2a^2 = 0

Solving for b using the quadratic formula yields:

b = a, which can be discarded
b = -2a

Since b = -2a:
f'(b) = 3b^2
f'(b) = 3(-2a)^2
f'(b) = 12a^2

And since f'(a) = 3a^2, the slope of the tangent at Q is four times the slope of the tangent at P.

Correct?

 

[Simon Bridge]

There is more than one way to skin a cat - one way of checking your results is to try a different way:

Brute force method:
1. find the slope of tangent at P
2. from that, find the equation (y_p) of the line tangent to y at P
3. find Q as the intersection of y_p and y.
4. find slope of tangent at Q using result in 3.

Let points P=(p,p^3) and Q=(q,q^3) and y=x^3 (1)
(This way there is no confusion about which belongs to what.)

y^\prime(p)=3p^2 (2)

Thus: the tangent line to P is y_p=(3p^2)x-2p^3 (3)

Intersection of (1) and (3) means y=y_p - gives:

x^3-(3p^2)x+2p^3=0 (4)

Roots of a cubic... (Someone more talented than me can do this by inspection.)
In standard form: a=1, b=0, c=-3p², d=2p³
The discriminant is: \Delta= 0 so there are multiple real roots.

In fact there are exactly 2:x_1= -2p ; x_2=x_3=p (5)

... so q=-2p

slope at Q is y^\prime(q)=3q^2 = 3(-2p)^2 = (4)3p^2 = 4y^\prime(p) (from (2))

Thus the slope at Q is 4x the slope at P.
[that q=-2p is the same as your b=-2a from a different path]
------------------------

Your strategy is a bit tricky to follow - let's see if I can tidy it up a bit:

The slope of the line between P and Q is given by:

\frac{\Delta y}{\Delta x} = \frac{p^3-q^3}{p-q}=3p^2 ...(6)
(last term from equation (2))

[putting it this way makes your reasoning obvious]

observe: p=q+(p-q)=q+h, so the slope is:
3p^2=\frac{(q+h)^3-q^3}{h}

(strategy is to use this formula to express q in terms of p)
[making strategy explicit]

[jumping ahead]

quadratic equation ... [write it out explicitly]:

q= \frac{-p \pm \sqrt{ p^2-4(1)(-2p^2) }}{2}=\frac{-p \pm 3p }{2}

So q={p,-2p}

[show you understand what it means]

i.e. there are two points of intersection - one at x=p, which is where we started, and the other at x=q=-2p

The rest follows the same.

Your reasoning seems OK - if your expression is not what I'm used to.

-------------------------------
what was the method in the book?

 

[SammyS]

I'm a beginning calculus student and this is one of the 'challenge' questions from an older edition of the Stewart's Calculus series of textbooks.
...

The Attempt at a Solution

Let the x-coordinate of P be "a", and let the x-coordinate of Q be "b". Also, let the slope of the tangent at P be "mL", and let the slope of the tangent at Q be "mK".

1. First I considered the tangent at P to be a secant line since it intersects the graph at two points. Using the formula for the slope of the secant, (f(x + h) - f(x))/h, I got:

(f(b + h) - f(b)) / h = mL
3b^2 + 3bh + h^2 = mL
...

Instead of introducing h (although there's nothing wrong with that) simply write the slope of the secant line as \displaystyle m_L=\frac{f(b)-f(a)}{b-a}\,. There is no need to consider the relative location of a & b on the number line.

Then \displaystyle m_L=\frac{b^3-a^3}{b-a}

\displaystyle =\frac{(b-a)(b^2+ab+a^2)}{b-a}

\displaystyle =b^2+ab+a^2​

etc.

 

[Seinfeld4]

Thank you both for the very detailed responses.

@Simon Bridge:

Your 'brute force method' is actually exactly how the book does it. I suppose that must be the most direct and least convoluted way to do it. I have a bad habit of making math problems seem more complicated than they really are!

Thanks again guys.

 

[Simon Bridge]

No worries - I think yours is more elegant :) since you don't need the horrible cubic root thing ... but it needs to be more clearly described. Brute force approach is charmingly naive and easy to follow - and finding the roots is hugely simplified by a=1 and b=0. More than half the terms vanish. That formula is not something you'd be expected to memorize for exams.

SammyS is correct that you didn't need to use that "h" - since you are putting it back later. I suspect you used a formula you looked up?

This is why I took some time over it - it looked like you'd stumbled upon a method without quite knowing what you were doing ... aside: about the secant...
If you draw a circle radius 1, pick a point P on the circumference then draw in the radius OP, now draw the tangent to the circle at P (this has a point - draw the picture), now pick another point Q in the same quadrant as P on the circumference - draw the line from O through Q until it intersects the tangent line, call the intersection T.

Observe that TOP forms an angle A.

The length of the line segment OP is 1 (the radius).
The length of the line segment PT is the tangent of A
The length of the line segment OT is the secant of A

(The length of the arc between P and Q is the size of A in radiens.)

Cool huh?

Back to your problem - you can make a similar triangle - the secant line is from P to Q, the tangent line is the tangent to the curve at point Q, so there is a "radius line" perpendicular to the tangent line, that passes through P. Is there a relationship between this radius and the radius of curvature? (Probably nothing special. Buggered if I know I only just thought of it!)

 

[Seinfeld4]

That's a very interesting geometric interpretation. I tried to solve this question again using a trigonometric approach, but conceded after about 30 minutes of not really getting anywhere.

It's actually interesting that the solution in the book does not require the use of the complicated formula for cube roots. Instead, it relies on the fact that (x - p) must be a root of the equation since this is the point of intersection of y = x^3 and the tangent at P. From there, it proceeds to use long division to find the factor (x + 2p).