How Fast is the Shadow Moving on Building B?

  • Context: MHB 
  • Thread starter Thread starter ISITIEIW
  • Start date Start date
  • Tags Tags
    Change
Click For Summary

Discussion Overview

The discussion revolves around a problem involving a tightrope walker crossing between two buildings, specifically focusing on calculating the speed of the shadow cast by the walker on the ground and on the wall of the adjacent building. The problem includes elements of related rates in calculus and requires understanding of geometric relationships and differentiation.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the initial problem setup and seeks help with part b of the question regarding the shadow's speed on the wall of building B.
  • Another participant agrees with the initial approach and suggests using implicit differentiation while noting the need to apply the product rule.
  • A participant calculates the rate of change of the shadow's height on building B, arriving at a value of -25/6 ft/s, using specific values for distance and rate of the tightrope walker.
  • Another participant proposes an alternative expression for b as a function of x, leading to a different formulation of the derivative.
  • There is a disagreement regarding the coefficient used in the calculations, with one participant asserting that the correct value is 4800, while another suggests it should be 480, leading to confusion and a correction later acknowledged by the first participant.

Areas of Agreement / Disagreement

Participants express differing views on the correct coefficient in the calculations, indicating a lack of consensus on this point. The discussion remains unresolved regarding the correct interpretation of the mathematical expressions involved.

Contextual Notes

The discussion includes potential misunderstandings related to the differentiation process and the values used in the calculations, which may depend on how the relationships are defined. There are also issues with the clarity of the visual aid provided, which may affect the understanding of the problem.

ISITIEIW
Messages
17
Reaction score
0
A Tightrope is 40ft above ground between two buildings that are 60 feet apart. A tightrope walker starts along the rope and walks from building A to building B at a rate of 2 ft per second. 80 feet above the starting point of the tightrope walker on building A is a spotlight that is illuminating the tightrope walker as the tightrope walker is crossing between two buildings.
a)How fast is the shadow of the tightrope walker's feet moving along the ground when the tightrope walker is midway between the buildings?
b)How fast is the shadow of the tightrope walker's feet moving up the wall of building B when the tightrope walker is twelve feet away from building B?

OK ! :)
I posted a picture of what I've done so far…
I got the answer for a) but I'm having troubles with b) as you can see in the picture.. Thanks
View attachment 1623
 

Attachments

  • math.JPG
    math.JPG
    26.2 KB · Views: 172
Physics news on Phys.org
The image you posted is sideways and is hard to read even when I enlarge and rotate it, but sometimes images posted from a mobile device display correctly on the device but not for those of us still using PCs. Interestingly, this problem was posted by another user earlier today, and your IP addresses are very close. :D

Anyway...I agree with what you have:

$$\frac{80}{x}=\frac{b}{60-x}$$

You then cross-multiplied

$$bx=80(60-x)=480-80x$$

I would next implicitly differentiate with respect to time $t$, bearing in mind that $b$ is a function of $t$ and so you need to use the product rule on the left side. What do you find?
 
Thanks,

I got b=4800/x - 80

then db/dt = -4800x^-2 (dx/dt)

and we know the values of x which equals 60-12=48 and (dx/dt)= 2

so db/dt = -25/6 ft/s

Thanks !
 
ISITIEIW said:
Thanks,

I got b=4800/x - 80

then db/dt = -4800x^-2 (dx/dt)

and we know the values of x which equals 60-12=48 and (dx/dt)= 2

so db/dt = -25/6 ft/s

Thanks !

If you choose (which is actually simpler) to write $b$ as a function of $x$, then you have:

$$b=480x^{-1}-80$$

and so:

$$\frac{db}{dt}=-480x^{-2}\frac{dx}{dt}$$

Your result is too large by a factor of 10, since you used 4800 instead of 480.
 
Um, not too sure that 480 is correct because bx= 80(60-x)
and 80 times 60 is 4800 not 480
Thanks again! :p
 
ISITIEIW said:
Um, not too sure that 480 is correct because bx= 80(60-x)
and 80 times 60 is 4800 not 480
Thanks again! :p

(Doh) You are absolutely right! Sorry for the confusion. I guess I am having one of those days...(Rofl)

I will go to the corner for 5 minutes. (Emo)
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 8 ·
Replies
8
Views
4K