Tangent to Graph of y = e^x

Date: 10/09/2003 at 15:27:34
From: Umair
Subject: graph of e^x

Hi Dr. Math.  I've got this problem where on the graph of y = e^x lies
a point Q with coordinates (b,a).  The tangent to Q makes an angle x
with the x-axis.  Why is it that tan(x) = a?

Date: 10/11/2003 at 11:44:24
From: Doctor Luis
Subject: Re: graph of e^x

Hi Umair,

At the point (b,a), the slope of the tangent to the curve y = e^x is
given by evaluating the derivative y'(b).  Since y' = e^x, we have
y'(b) = e^b.  However, the point (b,a) is ON the curve!  Therefore 
a = e^b.

Since tan(alpha) = m, and m = y'(b) = e^b = a, it follows that 
tan(alpha) = a.

To picture this geometrically, picture the curve y = e^x, the point
(b,a), and its tangent at that point: y = a*(x-b) + a

   

Now that you can visualize the triangle with the angle alpha, the
height is very clearly 'a'.  The base is a bit tricky to obtain.  The
intersection of the tangent line on the x-axis is obtained from
0 = a*(x-b)+a, or x = b-1.  Since the whole length from the origin is
'b', it follows that the base of the triangle is simply b-(b-1) = 1.

Therefore, tan(alpha) = a/1 = a.

I hope this helped!  Let us know if you have any more questions.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/