Curve and tangent line problem, finding the area of enclosed region

[needingtoknow]

Homework Statement

Given that the curve y = x^3 has a tangent line that passes through point (0, 2), find the area of the region enclosed by the curve and the line by the following steps.

Homework Equations

The Attempt at a Solution

Let f(x) = x^3 and let the coordinates of the point of tangency be (t, t^3)
f'(x) = 3x^2, f'(t) = 3t^2
The equation of the tangent line is:
y = 3t^2(x-t)+t^3
= 3t^2x - 2t^3

I am having trouble understand this part of the process. Particularly this step, y = 3t^2(x-t) + t^3. I have no idea what is happening at this step. Everything else I am completely aware of what is happening. If someone could help that would be great. Thanks!

 

[janhaa]

1) try to express x with respect to t, thus: y = 3t^2(x-t)+t^3 = x^3, further x = -2t

2) put x = -2t into the tangent line and use (0, 2), then you get; y - 2 = f ' (-2t)*(x - 0) = 3*(-2t)^2*x

then; y = 12t^2*x + 2

3) then: y = 12t^2*x + 2 = x^3
for a = 0,5 and x = -1
and
for a = 0,5 and x = 2

finally you can find the area...

 

[haruspex]

The method works like this:
Suppose the tangent line touches the curve at the point (t, t³). You wish to find t.
At that point, the gradient of the curve will be 3t², so that is the gradient of the tangent line. Hence you can write the equation of the tangent line as y = (3t²)x + c.
This line must also pass through the point (t, t³). Substituting that for (x, y) gives you an expression for c:

y = 3t²x - 2t³

Finally, you need that to be the equation of a line that passes through the point (0, 2). What value of t achieves that?

 

[needingtoknow]

Oh I see it makes so much more sense now! Thank you