1. ## Calculas!

I have another question to post here.
Write the equation of the straight line passing through the point (2,1) and normal to the curve 4y=x^2.

I know one thing x=2 and y=1 but I have no idea where to start or to do anything. How does the equation shown derivate.

Thanks

2. Originally Posted by Awsom Guy
I have another question to post here.
Write the equation of the straight line passing through the point (2,1) and normal to the curve 4y=x^2.

I know one thing x=2 and y=1 but I have no idea where to start or to do anything. How does the equation shown derivate.

Thanks
The curve is $\displaystyle 4y = x^2$

or $\displaystyle y = \frac{1}{4}x^2$.

It is easy to see that $\displaystyle (2, 1)$ lies on this curve.

$\displaystyle \frac{dy}{dx} = \frac{1}{2}x$

and when $\displaystyle x = 2, \frac{dy}{dx} = 1$.

So the tangent will have gradient = 1, which means the normal will have gradient $\displaystyle -\frac{1}{1} = -1$.

So the normal has gradient -1 and passes through (2, 1)

$\displaystyle y = mx + c$

$\displaystyle 1 = -1(2) + c$

$\displaystyle 1 = -2 + c$

$\displaystyle c = 3$.

So the equation of the normal is $\displaystyle y = -x + 3$.

3. ## sorry one more little thing...

one little question is the derivative of 2 equal to 1.

4. Originally Posted by Awsom Guy
one little question is the derivative of 2 equal to 1.

no, $\displaystyle \frac{d}{dx}(2)= 0$

5. Then why is the above derivative of 2 = 1. explain please.

6. The derivative is $\displaystyle \frac{1}{2}x$.

What does this equal when $\displaystyle x = 2$?

7. I think you are confsued, Prove it has said if

$\displaystyle y = \frac{1}{4}x^2$ then $\displaystyle \frac{dy}{dx} = \frac{1}{2}x$

This is the only derivate that has been taken.

The following is not a derivative, Prove it is saying at the point $\displaystyle x = 2,$ then $\displaystyle \frac{dy}{dx} = 1$

I.e from the first statement

$\displaystyle \frac{dy}{dx} = \frac{1}{2}x$

Put $\displaystyle x = 2$ here gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}\times 2 =1$

8. Right thanks for explaining that. I appreciate that.

9. The derivative of $\displaystyle \frac{1}{4}x^2$, $\displaystyle \frac{1}{2}x$ at x= 2 is 1. The "derivative of 2" means the derivative of the constant function "f(x)= 2" which is 0 because the derivative measures the rate of change and a constant function doesn't change.