Related Rates Word Problem :33

[StarlitxSunshine]

Here is the question & below are my steps. I have the answer, but I cannot seem to get to it.

A light shines from the top of a pole 50 ft high. A Ball is dropped from the same height from a paoint 30 ft away from teh light as shown below. How fast is the ball's shadow moving along the ground 1/2 sec later?
(Assume the ball falls a distance s=16t^2 in t sec)

dx/dt = ?
s= 16^2
t= 1/2

I set up a proportion as the equation:
50/50-s = x/x-30

50x-1500 = 50x -sx

Then differentiated to find related rates:

50dx/dt = 50 dx/dt - sdx/dt

But here, even with simplication, I end up with:

sdxt/dt = 0

-16(1/2)^2 * dx/dt = 0

dx/dt will equal zero. When the answer is supposed to be -150 m/s

What am I doing wrong, please ?

 

[Deleted member 4993]

Here is the question & below are my steps. I have the answer, but I cannot seem to get to it.

A light shines from the top of a pole 50 ft high. A Ball is dropped from the same height from a paoint 30 ft away from teh light as shown below. How fast is the ball's shadow moving along the ground 1/2 sec later?
(Assume the ball falls a distance s=16t^2 in t sec)

dx/dt = ?
s= 16t^2 = 4
t= 1/2

I set up a proportion as the equation:
50/50-s = x/x-30

50x-1500 = 50x -sx

Then differentiated to find related rates:

50dx/dt = 50 dx/dt - sdx/dt

But here, even with simplication, I end up with:

sdxt/dt = 0 <<< Here is your problem - s is a function of 't', not a constant

sx = 1500 <<<< at S = 4, x = 375

s * dx/dt + x * ds/dt = 0 <<<< Now continue...

-16(1/2)^2 * dx/dt = 0

dx/dt will equal zero. When the answer is supposed to be -150 m/s

What am I doing wrong, please ?

 

[wjm11]

A light shines from the top of a pole 50 ft high. A Ball is dropped from the same height from a paoint 30 ft away from teh light as shown below. How fast is the ball's shadow moving along the ground 1/2 sec later?
(Assume the ball falls a distance s=16t^2 in t sec)

dx/dt = ?
s= 16^2
t= 1/2

I set up a proportion as the equation:
50/50-s = x/x-30

50x-1500 = 50x -sx

Then differentiated to find related rates:

50dx/dt = 50 dx/dt - sdx/dt

(Note: Subhotosh, you were too quick for me! I have held off a little while now, but wanted to reward this student for supplying such a nice sketch of the problem! So, with apologies for my intrusion...)

Your differentiation (as Subhotosh pointed out) is in error.

50x-1500 = 50x –sx
1500 = sx
1500(s)^(-1) = x
[-1500(s)^(-2)](ds/dt) = dx/dt

Some geometry/physics in the problem: At t = .5s, s = 16t^2 = 16(.5)^2 = 4 ft.
Additionally, since s = 16t^2, ds/dt = 32t; at t = .5s, ds/dt = 16 ft/s.
Therefore,

dx/dt = [-1500(s)^(-2)](ds/dt) = [(-1500 ft^2)(4 ft)^(-2)](16 ft/s) = -1500 ft/s (NOT m/s)

Some comments about this problem: In general, I’d recommend using “y” for the vertical axis, and I would define “downward motion/displacement” as negative – not positive as it is defined here. It makes more sense “intuitively” if dy/dt and dx/dt are increasing or decreasing together in this problem.

You may also notice that I inserted units into the final math equation. Those units could have appeared throughout the entire solution. In fact, doing so (keeping units in every step) is a good check on our work; if the units don’t come out properly in the end, we’ve made a mistake somewhere along the way. And if our time units are in seconds, that’s a good reason not to use “s” to represent distance.

Please check my work for accuracy.