Calculus Q... PLEASE HELP!!!!

[tea217]

Picture

In the accompanying diagram, a person at point P is walking from A to B at 8m/s. His shadow is cast from point L to point S on the circular wall. Calculate the maximum speed of the person's shadow. Radius is 20m.


I spend so much time on this damn question...and i can't solve it...><

 

[roncarter]

the answer is 2 miles

hahahahahaha

 

[goodoptics]

Let me take a guess,

16m/s ???

 

[JonBarillari]

Define theta as angle OLA in radians, t as time in seconds, v as speed of person (8 m/s) and r as radius of circle (20 m)

Now (don't substitue for v yet): tan theta = (AO - vt) / r = (20 - vt) / 20 = 1 - vt/20

Differentiate implicitly with respect to time (A): sec^2 theta * d(theta)/dt = - (dv/dt t/20 + v/20)

Conceptually (I'll prove later if you need help) the maximum speed of the person's shadow occurs when person is at: O thus: theta = 0, dv/dt = d(8)/dt = 0 and t = 20/v = 5/2

Now substitute all values into (A): sec^2 (0) * d(theta)/dt = - [ 0 * ( (5/2) /20) + (8) / 20 ] = - 2/5 = d(theta)/dt Which means the rate of change of angle OLA is REDUCING at the SPEED of 2/5 randians per second

Define LS (Light to shadow) as the rotating arm with L the origin, thus: The speed at point S (the shadow) if the angular speed multiplied by the length of rotating arm LS (remeber, LS is found when P is at O so LS=2r=40m).

Thus, the maximum speed of the shadow is d(theta)/dt * LS = (2/5) * (40) = 16 m/s which was goodoptics guess! 😀 (All that work for nothing!!! 😱 o well)

Hope this helps!

 

[Kadarin]

<< Hope this helps! >>



Jon: That has to be the best first post ever! Welcome to the forums.. 😀

 

[JonBarillari]

<< Jon: That has to be the best first post ever! Welcome to the forums.. 😀 >>



Thanks 🙂

 

[tea217]

JonBarillari
thx...

Can u prove to me why the max speed occurs at O??
And..



<< Now substitute all values into (A): sec^2 (0) * d(theta)/dt = - [ 0 * ( (5/2) /20) + (8) / 20 ] = - 2/5 = d(theta)/dt Which means the rate of change of angle OLA is REDUCING at the SPEED of 2/5 randians per second >>



why don't use substitue d(0)/dt that?? "sec^2 (0) * d(theta)/dt " that part of the equation..

Also.. why does d(theta)/dt gives the RATE OF CHANGE? can u elaborate

 

[goodoptics]

<< Define theta as angle OLA in radians, t as time in seconds, v as speed of person (8 m/s) and r as radius of circle (20 m)

Now (don't substitue for v yet): tan theta = (AO - vt) / r = (20 - vt) / 20 = 1 - vt/20

Differentiate implicitly with respect to time (A): sec^2 theta * d(theta)/dt = - (dv/dt t/20 + v/20)

Conceptually (I'll prove later if you need help) the maximum speed of the person's shadow occurs when person is at: O thus: theta = 0, dv/dt = d(8)/dt = 0 and t = 20/v = 5/2

Now substitute all values into (A): sec^2 (0) * d(theta)/dt = - [ 0 * ( (5/2) /20) + (8) / 20 ] = - 2/5 = d(theta)/dt Which means the rate of change of angle OLA is REDUCING at the SPEED of 2/5 randians per second

Define LS (Light to shadow) as the rotating arm with L the origin, thus: The speed at point S (the shadow) if the angular speed multiplied by the length of rotating arm LS (remeber, LS is found when P is at O so LS=2r=40m).

Thus, the maximum speed of the shadow is d(theta)/dt * LS = (2/5) * (40) = 16 m/s which was goodoptics guess! 😀 (All that work for nothing!!! 😱 o well)

Hope this helps! >>



Now the question is:
How I came up with that guess?

Well, it's been a long time since I've taken calculus. I used vectors and similar triangle and conceptually, max velocity of the shadow occurs when the person is at point O. I further argued that if the person is walking 8 m/s to the right, when the person is at point O, the shadow must travel at an instantaneous velocity of 16 m/s to the right in order to keep up with the person.

 

[Fandu]

<< Also.. why does d(theta)/dt gives the RATE OF CHANGE? can u elaborate >>



That's what the first derivative is, say it as "the rate of change of theta with respect to t"

If you think of the first derivative as the tangent line to a curve, then the derivative at any point on the curve is the slope of the curve at that exact point, and only that exact point. And we can call the slope "the rate of change of y with respect to x"

 

[tea217]

<< If you think of the first derivative as the tangent line to a curve, then the derivative at any point on the curve is the slope of the curve at that exact point, and only that exact point. And we can call the slope "the rate of change of y with respect to x" >>


Well..i kinda of get this statement.
So, if dy/dx = 8, this is saying "the slope is changing at the rate of 8m/s"???

Also, can anyone tell me why when the light hit the center of the circle, this gives the max speed of shadow???? pls tell me.......
i appreciate all ur help...

 

[Fandu]

<<

<< If you think of the first derivative as the tangent line to a curve, then the derivative at any point on the curve is the slope of the curve at that exact point, and only that exact point. And we can call the slope "the rate of change of y with respect to x" >>


Well..i kinda of get this statement.
So, if dy/dx = 8, this is saying "the slope is changing at the rate of 8m/s"???

Also, can anyone tell me why when the light hit the center of the circle, this gives the max speed of shadow???? pls tell me.......
i appreciate all ur help... >>




No, the slope is not changing, y is changing at a rate of 8m/sec with respect to x.
Derivatives are not something that you can apply to an entire curve, every point on the curve has a different derivative. So at the exact point where you take the derivative, and only at that point, the slope is 8. Slope is rise/run, or x/y on a 2d coord system.

Think about it, as the shadow moves farther from the person, it's speed must increase. If you want to see this graphically, take your derivative function and graph it. You'll see a smooth curve that will come up, peak, and then drop off again. It will peak right at (B-A)/2, or 'O'

 

[JonBarillari]

<< Also, can anyone tell me why when the light hit the center of the circle, this gives the max speed of shadow???? pls tell me....... >>



The speed of the shadow is given by: d(theta)/dt * LS (rotating arm length times angular (rotating) speed) so you could find an equation for the chord LS and multiply it by our equation for d(theta)/dt (before subing in values) then do traditional maximizing calculus methods, but, the chord length is beyond my ability 🙁

Another way: luckily for this question, because both angular speed and rotating arm length are maximized at the same point (theta = 0) (LS's maximum is the diameter and d(theta)/dt is maximzed when sec (theta) is maximized (from equation)) this implies shadow speed is maximized at the same point.

Hope this helps!

 

[tea217]

thank you everyone...i UNDERSTAND IT
😀