# Dynamic Equation of the Tangent Line Problem

• Oct 21st 2009, 12:07 PM
NBrunk
Dynamic Equation of the Tangent Line Problem

My math professor likes to give us problems that require a lot of synthesizing the various things we've learned recently with things we should already know. She stated this when working out the solution to this problem, but I still don't fully understand it. The problem is this:

Find the equation of the tangent line to f(x) = e^x that passes through the origin.

She stressed the difference between finding the tangent line to the function at x = 0 and finding the tangent line that passes through both x & y = 0.

Any assistance someone could give in clearing up this tricky problem would be greatly appreciated.
• Oct 21st 2009, 12:58 PM
Quote:

Originally Posted by NBrunk

My math professor likes to give us problems that require a lot of synthesizing the various things we've learned recently with things we should already know. She stated this when working out the solution to this problem, but I still don't fully understand it. The problem is this:

Find the equation of the tangent line to f(x) = e^x that passes through the origin.

She stressed the difference between finding the tangent line to the function at x = 0 and finding the tangent line that passes through both x & y = 0.

Any assistance someone could give in clearing up this tricky problem would be greatly appreciated.

One thing that we know for sure is that the tangent will touch the curve at some point in the 1st quadrant of the xy plane. Do you see why? We also know that (1,1) is a point on the curve. So there is a point (x,y) through which the tangent passes, there is a point (0,0) through which the tangent passes. So (1,1),(x,y),and (0,0) form a triangle. You should be able to do something with that.

I'm trying to work it out on paper, but I think that is a good place to start.
• Oct 21st 2009, 01:43 PM
Tangent Lines to Exponential and Logarithmic Functions through the Origin - Wolfram Demonstrations Project

It seems that point of tangency will be at y=e according to the source above. It's not clear to me why this is true though. But it's useful information.
• Oct 21st 2009, 02:43 PM
Plato
Quote:

Originally Posted by NBrunk
Find the equation of the tangent line to f(x) = e^x that passes through the origin.

Any tangent line to \$\displaystyle f(x) = e^x \$that passes through the point \$\displaystyle (a,e^a)\$ looks like \$\displaystyle y=e^a(x-a)+e^a\$.
So the origin is to be on that line we would have be \$\displaystyle 0=e^a(0-a)+e^a\$.
But that means that a=1.
Why are we sure that \$\displaystyle a\ne 0\$?