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)

Printable View

- Jan 31st 2013, 06:18 PMmathisfun26Finding the vaue of c given an equation of line
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) - Jan 31st 2013, 08:43 PMchiroRe: Finding the vaue of c given an equation of line
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). - Jan 31st 2013, 08:48 PMabenderRe: Finding the vaue of c given an equation of line
$\displaystyle f(x)=c\sqrt{x}$

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

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

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

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