# [SOLVED] Differentiation Help!

• Mar 29th 2008, 01:28 PM
looi76
[SOLVED] Differentiation Help!
Question:Find the value of the constant c for which the line $y = 2x+c$ is a tangent to the curve $y^2 = 4x$

Attempt:
NO IDEA (Crying)
• Mar 29th 2008, 01:44 PM
wingless
Hint: $y=2x+c$ tells us the tangent lines slope is $m=2$. So y' is 2 at the tangent point.
• Mar 29th 2008, 01:46 PM
TrevorP
Well think about it. You are told that you are to find a line that is TANGENT to another line.

This means when the derivative of the second equation equals the slope of the first one:

$y^2 = 4x \rightarrow y = 2\sqrt{x}$

and you want when it is equal to 2:

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

AND also you know that they both are equal at those x values. So you should be able to get two equations and solve them for C.
• Mar 29th 2008, 02:50 PM
wingless
Quote:

Originally Posted by TrevorP
Well think about it. You are told that you are to find a line that is TANGENT to another line.

This means when the derivative of the second equation equals the slope of the first one:

$y^2 = 4x \rightarrow y = 2\sqrt{x}$

Not true.
Remember: $\sqrt{x^2}$ is NOT $x$, its $|x|$!

So,
$y^2 = 4x$

$\sqrt{y^2} =\sqrt{4x}$

$|y| = \sqrt{4x} = 2\sqrt{x}$

This gives us two functions of y:
$y= 2\sqrt{x}$ and $y= -2\sqrt{4x}$.

Now differentiate both and set both derivatives y' = 2. Only one of them will satisfy it and you'll get one value for x. This is the x coordinate of the tangent point. Find its y coordinate too and you'll get the tangent point. Plug this point in the line equation to get c.

A better approach is here:

If $\frac{dy}{dx} = 2$, then $\frac{dx}{dy} = \frac{1}{2}$

$y^2 = 4x$

$x' = \frac{y}{2}$

Now set $x'=2$ to get the tangent points y coordinate. This is much easier than working with two functions.