# Tangent to Given Curve

• Oct 7th 2010, 06:46 PM
stiitches
Tangent to Given Curve
Need help on this problem

http://imgur.com/0KPop.jpg

have 0 idea what to do/how to start it... (Worried)
• Oct 7th 2010, 07:03 PM
Jhevon
Quote:

Originally Posted by stiitches
Need help on this problem

http://imgur.com/0KPop.jpg

have 0 idea what to do/how to start it... (Worried)

tangent to the given curve at what point?

anyway, remember, the derivative gives the slope of the tangent line at any point.
• Oct 7th 2010, 07:05 PM
stiitches
Quote:

Originally Posted by Jhevon
tangent to the given curve at what point?

anyway, remember, the derivative gives the slope of the tangent line at any point.

I dont know, thats the whole question...
So I should start by finding the derivative of the first function? or the second one?
• Oct 8th 2010, 06:40 AM
HallsofIvy
Quote:

Originally Posted by stiitches
I dont know, thats the whole question...
So I should start by finding the derivative of the first function? or the second one?

In order for a line to be tangent to a curve, at a given point, both curve and line must pass through the point and must have the same derivative there.

So you need to find c and x such that $y= \frac{5}{6}x+ \frac{15}{2}= c\sqrt{x}$ and they have the same derivative for that x. That is two equations to solve for x and c.

I will give you one derivative free- the derivative of $y= \frac{5}{6}x+ \frac{15}{2}$ is $\frac{5}{6}$, for all x. What is the derivative of $c\sqrt{x}$? (That will depend on x.)