<p>Ok, so here's the question:</p>
<p>A man 2 meters tall walks at a rate of 2 meters per second towards a streetlight that's 5 meters tall. At what rate is the tip of the man's shadow moving?</p>
<p>i have no idea how to do it.</p>
<p>Sun.
Draw a line tangent to the man's head and the shadow.
Similar triangles.
Derivate.
Cheer.</p>
<p>oh yeah that's right. thanks</p>
<p>I got 10/3 and if it's right here's how I did it.
Draw a vertical line (the streetlight) and label it 5. Make a horizontal line for the ground and then connect the top of the streetlight to the end of the ground to make a triangle. Pick an arbitrary value for the length of the ground, say 4. Then make a vertical line to the ground inside the triangle to represent the man (with a height of 2). The whole triangle and the interior triangle you just created are similar. The line from the bottom of the interior triangle to the end of the ground represents the man's shadow. So, set up the proportion 5/2 = 4/x to find the current length of the shadow, where x is the shadow's length. X= 8/5. If the man were to walk all the way to the streetlight and stand directly under it, the shadow would lie on top of him. The man would have walked 4-(8/5) =12/5 meters. The shadow would have traveled the whole 4 meters. Since 4 is 166.666% of 12/5, that means the shadow traveled 1.666 times faster in the same amount of time. Thus multiply 1.666 times 2mph and you get 10/3 mph. Hopefully this is right lol</p>
<p>is this right??? now I just want to know for my sake lol</p>
<p>Yes, it's right. I used proportions, too, but I worked it a bit differently.</p>
<p>I drew a diagram with the man a distance x away from the lightpole. Using proportions, the man's shadow is 2x/3 and the distance from the tip of his shadow to the lightpole is 5x/3. Now, if dx/dt = 2 m/s, then d(5x/3)/dt = (5/3)*dx/dt = 10/3 m/s.</p>
<p>Yep. You got it. I'm not too big of fan of those Change of Rates problems. Some of them are pretty tough.</p>
<p>Best of Luck</p>
<p>Jerod</p>
<p>I did the same thing they did, but instead of setting the ground equal to something random, I set the distance the man is from the pole equal to x, and then the length of the shadow equal to y (so the distance from the shadow to the lightpole is x + y). Also, dx/dt, the rate the distance between the man and the poll is changing at, is -2 m/s. Now you can set up the proportion:</p>
<p>5/(x+y) = 2/y</p>
<p>And x and are y are thus related:</p>
<p>y = (2/3)x</p>
<p>dy/dt = (2/3)(dx/dt) = (2/3)(-2) = -4/3 m/s</p>
<p>That's the rate that the shadow is decreasing at, then you just add it to the rate that the man is walking towards the poll at:</p>
<p>-4/3 + -2 = -10/3 m/s</p>