1. ## find a vector

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

2. 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 $\displaystyle y = mx + c$, to find $\displaystyle m$, take the derivative, then evaluate the derivative at that point.

$\displaystyle y = x^2$

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

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

Thus $\displaystyle m = 4$.

We have

$\displaystyle y = 4x + c$

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

So $\displaystyle 4 = 4\cdot 2 + c$

$\displaystyle c = -4$.

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

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

Can you go from here?

3. 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 $\displaystyle y = mx + c$, to find $\displaystyle m$, take the derivative, then evaluate the derivative at that point.

$\displaystyle y = x^2$

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

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

Thus $\displaystyle m = 4$.

We have

$\displaystyle y = 4x + c$

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

So $\displaystyle 4 = 4\cdot 2 + c$

$\displaystyle c = -4$.

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

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

Can you go from here?