Elegant Problem (1 Viewer)

[OLDMAN]

Sorry for the long absence. offers a lame "Sydney" excuse: got sidetracked by real estate!

Two months till the big exam, I'd like to help. I'll pose an elegant problem now and then. What's an elegant problem? A problem charactererized by the brevity of its statement, giving nothing away. These problems are not necessarily difficult like the question that follows, but whereas the HSC examiner will lead you by the hand through parts i, ii, and/or iii, an elegant problem offers no such guideposts- thus an excellent practice question.

Question: Consider the curve y=x^3. The tangent at A meets the curve again at B. Prove that the gradient at B is 4 times the gradient at A.

 

[underthesun]

Long time no see

here's my solution ::

y = x³

we can have a parametric representation of A(t , t³), where x = t, and y = t³.

^dy/_dx = 3x² = 3t² = A_m

The tangent equation is :

y - t³ = 3t²(x - t)

this simplifies to

[ y = 3t²x - 3t³ ] <-- tangent eq

To find the intercepts with y = x³, we substitute the values of y.

x³ = 3t²x - 3t³

this simplifies to:

(let) p(x) = x³ - 3t²x + 2t³ = 0

however, as the tangent meets the curve y = x³ with a "stationary point" manner, p(x) should have a double root at x = t.

now, comparing roots
abc (the product of roots of p(x)) = -2t³.
at² = -2t³
a = -2t.

Hence,

p(x) = (x - t)²(x+2t) = 0

The tangent touches the graph again at x = -2t.

^dy/_dx = 3(-2t)² = 12t²

Hence, the gradient at B is 12t².

B_m/A_m=4

that's my solution. But is that how you are supposed to say that (red text).

nice question btw

 

[Affinity]

here's another solution, similar:

let A = (a,a^3) and B = (b,b^3) a != b (!= means not equal)
we know that gradient of AB is 3a^2 by differentiation
so:

(a^3 - b^3)/(a-b) = 3a^2
a^2 + ab+ b^2 = 3a^2 (since a-b is not 0)
b^2 +ab -2a^2 = 0
(b+2a)(b-a)= 0
b = -2a (again, since b-a !=0)
gradient at B is 3b^2 = 12a^2 which is 4 times that at A
Q.E.D.

 

[OLDMAN]

underthesun: first to cross the line, solid approach.
QUOTE "however, as the tangent meets the curve y = x3 with a "stationary point" manner, p(x) should have a double root at x = t." Or simply : point of tangency corresponds with repeated roots.
Affinity :What a creative alternative!