Tangent to Graph of y = e^xDate: 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/ |
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