Differential Approximation

[recoil33]

"Differential Approximation"

A person of height 1.9m is walking away from a lamp-post at 1m/s. The light on the lamp-post is 5.1m above the ground.


1. At what rate (in m/s to the nearest cm/s) is the length of the person's shadow changing?

2. At what speed (in m/s to the nearest cm/s) is the tip of the person's shadow moving across the ground?

Alright, I'll be one second to type my working out. I really do not understand what I'm doing much. Any help would be appreciated.



1.9/s = 5.1/s+x

ds/dt + dx/dt

1.9/s = 5.1/(s+x) ---> 1.9(s+x)=13s

1.9(ds/dt + dx/dt) = 13(ds/dt)

we know that dx/dt = 1 ms^-1 (the speed he's walking)

therefore.

s' + x' = (1.9*1)/(5.1-1.9) + 1 = 4.8/3.2

therefore, i got the shadow moving at 4.8/3.2 m/s which is wrong? ;s

 

[HallsofIvy]

A person of height 1.9m is walking away from a lamp-post at 1m/s. The light on the lamp-post is 5.1m above the ground.


1. At what rate (in m/s to the nearest cm/s) is the length of the person's shadow changing?

2. At what speed (in m/s to the nearest cm/s) is the tip of the person's shadow moving across the ground?

Alright, I'll be one second to type my working out. I really do not understand what I'm doing much. Any help would be appreciated.



1.9/s = 5.1/s+x

I presume you drew a picture with a horizontal line as the ground, the lamp post as one vertical leg and the person as another. Drawing a straight line from the lamp through the top of the person's head to the ground gives to similar right triangles. The height of the smaller triangle is the person's height, 1.9, and the base is the length of his shadow, s. The height of the larger triangle is the height of the lamp and the base is the person's distance from the lamp post plus the length of his shadow, x+ s.

(Suggestion- do not use symbols like "x" and "s" without explaining what they mean: "x is the distance from the lamp post to the person, in meters, and s is the length of his shadow, in meters.")

Since these triangles are similar, the ratios are the same: 1.9/s= 5.1/(s+ x) (notice the parentheses-- what you wrote would be properly interpreted as 1.9/s= (5.1/s)+ x). However, you could also write that proportion as s/1.9= (s+ x)/5.1 and I think you will find that easier.

ds/dt + dx/dt

Since this isn't equal to anything I have no idea what you mean or why you wrote this.

1.9/s = 5.1/(s+x) ---> 1.9(s+x)=13s

What? Where did the "13" come from? "Cross multiplying" gives 1.9(s+ x)= 5.1s
But somehow, that "13" gets changed back to "5.1" so it does not affect the result.

1.9(ds/dt + dx/dt) = 13(ds/dt)

we know that dx/dt = 1 ms^-1 (the speed he's walking)

therefore.

s' + x' = (1.9*1)/(5.1-1.9) + 1 = 4.8/3.2

therefore, i got the shadow moving at 4.8/3.2 m/s which is wrong? ;s

Differentiating, 1.9(s'+ x')= 5.1s'. and, since x'= 1, 1.9(s'+ 1)= 1.9s'+ 1.9= 5.1s'.

(1.9- 5.1)s'= -1.9 so that s'= 1.9/(5.1- 1.9)= 3.2 m/s which is what you appear to get. Since s is the length of the shadow, that value is the rate at which the length of the shadow is growing, the solution to the problem.

Adding 1 to s' gives the rate at which the tip of the shadow is moving away from the lam post which was not asked.