# Finding k of a Tangent Line

• Feb 4th 2010, 04:45 PM
BeSweeet
Finding k of a Tangent Line
http://img29.imageshack.us/img29/1954/capturend.png

Maybe we can walk through this? At first glance, the first thing I would do is find the derivative of $y=5x+45$, but that's not right.
• Feb 4th 2010, 04:50 PM
pickslides
Quote:

Originally Posted by BeSweeet
At first glance, the first thing I would do is find the derivative of $y=5x+45$, but that's not right.

first of all set the derivative - $(k\sqrt{x})' = 5$ and solve for x.
• Feb 4th 2010, 04:57 PM
BeSweeet
Quote:

Originally Posted by pickslides
first of all set the derivative - $(k\sqrt{x})' = 5$ and solve for x.

I'm still pretty lost :\.
• Feb 4th 2010, 05:03 PM
pickslides
Quote:

Originally Posted by BeSweeet
I'm still pretty lost :\.

$y = k\sqrt{x}$ find $\frac{dy}{dx}$

Then make $\frac{dy}{dx}=5$ as that is the gradient of the tangent.

Now solve for $x$
• Feb 4th 2010, 05:08 PM
BeSweeet
Quote:

Originally Posted by pickslides
$y = k\sqrt{x}$ find $\frac{dy}{dx}$

Then make $\frac{dy}{dx}=5$ as that is the gradient of the tangent.

Now solve for $x$

That's one of the things... We never learned the whole $\frac{dy}{dx}$ thing. We just used:
$(f(x+h)-f(x))/h$

...
• Feb 5th 2010, 03:46 AM
HallsofIvy
If you haven't learned "the whole dy/dx thing", why did you refer to finding the derivative in your first post? The formula you give is for the slope of a straight line. If f(x) is not linear, it is the slope of a "secant" line, a line that crosses the graph at x and x+y, not the tangent line. Since you don't yet know the derivative (I suspect this problem is preliminary to introducing the derivative), here's a method that predates the calculus:

If y= 5x+ 45 and $y= k\sqrt{x}$ meet at all, we must have $y= 5x+ 45= k\sqrt{x}$ or $5x- k\sqrt{x}+ 45= 0$. Let u= $\sqrt{x}$ so that $x= u^2$. The equation becomes $5u^2- ku+ 45= 0$. If x= a at the point of intersection then x- a= $u- \sqrt{a}$ must be a factor of that polynomia. If fact, to be tangent that must be a double factor- we must have $5u^2- ku+ 45= 5(u- \sqrt{a})^2$ Multiplying the right side, we get $5u^2- ku+ 45= 5u^2- 10\sqrt{a} u+ 5a$ for all u. Comparing coefficients, $-k= -10\sqrt{a}$ and $5a= 45$. Solve the second equation for a and use that to solve the first equation for k.