Let P be a point on the curve y=x^3 and suppose the tangent line at P intersects the curve again at Q...?

Question

Let P be a point on the curve y=x^3 and suppose the tangent line at P intersects the curve again at Q. Show that the slope at Q is 4 times the slope at P.

I'm kinda lost on this..
Thanks!

Answer 1

Amit might have this right, but to me it seemed hard to follow, mostly, perhaps, because of the (x0,y0) notation. I'm going to try this problem using different notation.

y = x^3
dy/dx = 3x^2

Let P be the point (a,a^3). The slope at P is 3a^2.

Using point-slope, the equation of the tangent line through P is

y - a^3 = (3a^2) (x - a)
y = (3a^2)x - 3a^3 + a^3 = (3a^2)x - 2a^3

The point Q is the intersection of y = x^3 and y = (3a^2)x - 2a^3. Since both of these equations are already solved for y, all we have to do is set the right sides equal and solve for x:

x^3 = (3a^2)x - 2a^3
x^3 - (3a^2)x + 2a^3 = 0

To simplify, we already know that x = a is a solution to this, so we can divide that out by polynomial long division:

(x - a) (x^2 + ax - 2a^2) = 0

Solve the quadratic by factoring:

x^2 + ax - 2a^2 = (x + 2a) (x - a) = 0

Interesting. There's a double root at x = a (point P), and the other root (point Q) is at x = -2a. Since the function is y = x^3, the coordinates of Q are (-2a, -8a^3). This is not an answer, but it's something we'd want to know anyway.

Our formula for the slope is dy/dx = 3x^2. At P, the slope is 3a^2, and at Q the slope is 3(-2a)^2 = 12a^2.

We see, therefore, that the slope at Q is four times the slope at P.

That completes your proof. I don't know if this makes any more sense than the one Amit already did. Take your choice.

Answer 2

Slope of the tangent line is equal to the derivative at a point. Differentiate using the product rule: y'(x) = 5 (-e^x sin(x) + e^x cos(x)) = 5e^x(cos(x) - sin(x)) Plug in the given point (x = 0): y'(0) = 5e^0(cos(0) - sin(0)) = 5(1)(1 - 0) = 5 Equation of a line: y = mx + b We just found m = 5: y = 5x + b Plug in the given point to find b: 5 = 5(0) + b b = 5 So the equation of the tangent line is: y = 5x + 5

Answer 3

To find the slope at any point derive the function:
y'=3x^2

Now, the tangent at P(x0,y0) is:

y=3(x0)^2 * x + (x0)^3-3(x0)^3 =
= 3(x0)^2 * x - 2(x0)^3 // So that y0 = x0^3

Now, let's find where it intersects with y=x^3

y=x^3
y=3(x0)^2 * x - 2(x0)^3

x^3 = 3(x0)^2 x - 2(x0)^3

x^3 - 3(x0)^2 x + 2(x0)^3 = 0

We know that x=x0 is one solution

x^3 - (x0)^2 *x - 2(x0)^2 x + 2(x0)^3 = 0

x[x^2 - (x0)^2] - 2(x0)^2 *(x - x0) = 0

x(x - x0)(x + x0) - 2(x0)^2 (x - x0) = 0

[x(x + x0) - 2(x0)^2](x - x0) = 0

[x^2 + x*x0 - 2(x0)^2](x - x0) = 0

Let's solve x^2 + x*x0 - 2(x0)^2 = 0

x1,2 = [-x0 +- sqrt((x0)^2 + 8(x0)^2]/2 =
= [ -x0 +- sqrt(9(x0)^2)]/2 =
= (-x0 +- 3x0)/2

x1 = x0 // we had it once
x2 = - 2x0

Plug x2=-2x0 in the derivative:

3(-2x0)^2 = 3*4*(x0)^2 = 4*3(x0)^2