The Speed of Change of a ShadowDate: 12/12/95 at 21:39:16 From: Anonymous Subject: Calculus, Rates A street lamp whose light is 5 meters from the ground is 6 meters from a vertical brick wall. A person 2 meters tall walks at right angles away from the wall directly towards the street lamp at a speed of 1.2 meters per second. At what speed is the shadow of the top of the person's head descending down the wall when the person is 1.5 meters from the wall. As with most of these problems the only difficulty I have occurs when I try to create an equation that will allow me for solve for dx/dt. I'm not sure how to start it but I tried using triangle laws to get some substitutions but nothing seemed to work. Thanks for your help. Date: 5/29/96 at 1:54:45 From: Doctor Anthony Subject: Re: Calculus, Rates Let x be the distance of the man from the wall, and y be the height of his shadow up the wall. Draw a horizontal line through the top of the man's head to give two similar triangles, which then allow us to write the following ratios as equal. 3/(6-x) = (2-y)/x Multiply this out to give 5x + 6y - xy = 12 Differentiate with respect to time 5(dx/dt)+6(dy/dt)-x(dy/dt)-y(dx/dt)=0 Now when x=1.5, the above ratio equation gives y=1. Also dx/dt = 1.2 Substitute these values the into the differential equation 5(1.2)+6(dy/dt)-1.5(dy/dt)-1.2 = 0 6 + 4.5(dy/dt) -1.2 = 0 4.5(dy/dt) = - 4.8 dy/dt = -4.8/4.5 = -1.06667 The negative value is correct since the shadow is moving down the wall and y is decreasing. -Doctor Anthony, The Math Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/