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10 Nov 2019
For the first part of the question, I got ....
ds/dt = (dh/dt (-Hs/h))/(h-H)
which was correct,..
but I don't know how to do the 2nd part of the question atall.
I know that it says that you can solve part 2 using limits, butI know that you can use a different method without computing alimit.
Help would really be appreciated.. I've been stuck on this forquite a while. ):
This is a related rates problem with a twist. Suppose you have a street light at a height H You drop a rock vertically so that it hits the ground at a distance d from the street light. Denote the height of the rock by h. The shadow of the rock moves along the ground. Let s denote the distance of the shadow from the point where the rock impacts the ground. Of course, s and h are both functions of time. To enter your answer into WeBWorK use the notation v to denote h': Then the speed of the shadow at any time while the rock is in the air is given by s' = (where s' is an expression depending on h, s, H. and v (You will find that d drops out of your calculation.) Now consider the time at which the rock hits the ground. At that time h = s = 0 The speed of the shadow at that time is s - where your answer is an expression depending on H, v, and d Hint: Use similar triangles and implicit differentiation. For the second part of the problem you will need to compute a limit.
For the first part of the question, I got ....
ds/dt = (dh/dt (-Hs/h))/(h-H)
which was correct,..
but I don't know how to do the 2nd part of the question atall.
I know that it says that you can solve part 2 using limits, butI know that you can use a different method without computing alimit.
Help would really be appreciated.. I've been stuck on this forquite a while. ):
This is a related rates problem with a twist. Suppose you have a street light at a height H You drop a rock vertically so that it hits the ground at a distance d from the street light. Denote the height of the rock by h. The shadow of the rock moves along the ground. Let s denote the distance of the shadow from the point where the rock impacts the ground. Of course, s and h are both functions of time. To enter your answer into WeBWorK use the notation v to denote h': Then the speed of the shadow at any time while the rock is in the air is given by s' = (where s' is an expression depending on h, s, H. and v (You will find that d drops out of your calculation.) Now consider the time at which the rock hits the ground. At that time h = s = 0 The speed of the shadow at that time is s - where your answer is an expression depending on H, v, and d Hint: Use similar triangles and implicit differentiation. For the second part of the problem you will need to compute a limit.