find a vector

**theliong** · September 25th 2009, 08:49 PM

how to find the vector that tangent to the parabola y=x^2 at (2,4) ??

**Prove It** · September 25th 2009, 09:31 PM

Originally Posted by theliong

how to find the vector that tangent to the parabola y=x^2 at (2,4) ??

It will be a 2D vector, and parallel to the line defined by the tangent at that point.

To find the tangent, which will be of the form $y = mx + c$ , to find $m$ , take the derivative, then evaluate the derivative at that point.

$y = x^2$

$\frac{dy}{dx} = 2x$

$\frac{dy}{dx}|_{x = 2} = 4$ .

Thus $m = 4$ .

We have

$y = 4x + c$

We also have $(x, y) = (2, 4)$ as a point on the tangent.

So $4 = 4\cdot 2 + c$

$c = -4$ .

Thus the tangent is $y = 4x - 4$ .

So you will have a vector that is parallel to $y = 4x - 4$ and that passes through $(x, y) = (2, 4)$ .

Can you go from here?

**theliong** · September 26th 2009, 05:59 AM

The answer is <1,4>. How to evaluate from y=4x-4 until <1,4>??

Originally Posted by Prove It

It will be a 2D vector, and parallel to the line defined by the tangent at that point.

To find the tangent, which will be of the form $y = mx + c$ , to find $m$ , take the derivative, then evaluate the derivative at that point.

$y = x^2$

$\frac{dy}{dx} = 2x$

$\frac{dy}{dx}|_{x = 2} = 4$ .

Thus $m = 4$ .

We have

$y = 4x + c$

We also have $(x, y) = (2, 4)$ as a point on the tangent.

So $4 = 4\cdot 2 + c$

$c = -4$ .

Thus the tangent is $y = 4x - 4$ .

So you will have a vector that is parallel to $y = 4x - 4$ and that passes through $(x, y) = (2, 4)$ .

Can you go from here?

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