Tangent to Given Curve

**stiitches** · October 7th 2010, 05:46 PM

Need help on this problem

have 0 idea what to do/how to start it...

**Jhevon** · October 7th 2010, 06:03 PM

Originally Posted by stiitches

Need help on this problem

have 0 idea what to do/how to start it...

tangent to the given curve at what point?

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

**stiitches** · October 7th 2010, 06:05 PM

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?

**HallsofIvy** · October 8th 2010, 05:40 AM

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.)

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