Finding the vaue of c given an equation of line

**mathisfun26** · January 31st 2013, 06:18 PM

Find the value of c such that the line y=5/6x+15/2 is tangent to the curve y= c(sqrt x)
I am not sure where to start please help
list after with finding the derivative of c(sqrt x)

**chiro** · January 31st 2013, 08:43 PM

Hey mathisfun26.

Hint: Start off by calculating the derivative (and noting that a tangent line has the form y = mx + b where m is the derivative).

**abender** · January 31st 2013, 08:48 PM

Originally Posted by mathisfun26

Find the value of c such that the line y=5/6x+15/2 is tangent to the curve y= c(sqrt x)
I am not sure where to start please help
list after with finding the derivative of c(sqrt x)

$f(x)=c\sqrt{x}$

$f^\prime(x)=\frac{c}{2\sqrt{x}}$

If $t(x)=\tfrac{5}{6}x + \tfrac{15}{2}$ is tangent to $f(x)$ , then the slope of the line tangent to $f(x)$ is $\tfrac{5}{6}$ .

To find the point at which $t(x)$ is tangent to $f(x)$ , we first solve for $x$ (in terms of $c$ ) in $\tfrac{c}{2\sqrt{x}}= \tfrac{5}{6}$ . Plug this $x$ -coordinate, which we label $x_0$ , into $f(x)$ to determine the $y$ -coordinate of the point at which $t(x)$ is tangent to $f(x)$ , which we call $y_0$ .

Finally, plug the point $(x_0, y_0)$ into $t(x)$ and you then should be able to solve for $c$ .

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