Asked by Nikki
(It's a related rate question)
A man 6ft tall walks away from a lamp post (15ft) at 5ft/sec. How fast is his shadow lengthening?
---
I have the picture, and the constants, the man and the lamp post. I have 5ft/sec as dw/dt, and I know I'm looking for ds/dt, the rate of the shadow.
Problem is, I don't know the actual equation to figure this out. Help please?
Thanks!
A man 6ft tall walks away from a lamp post (15ft) at 5ft/sec. How fast is his shadow lengthening?
---
I have the picture, and the constants, the man and the lamp post. I have 5ft/sec as dw/dt, and I know I'm looking for ds/dt, the rate of the shadow.
Problem is, I don't know the actual equation to figure this out. Help please?
Thanks!
Answers
Answered by
bobpursley
The lamp post makes a right triangle.
let x be the distance from the lamppost to the man (dx/dt=5; given) Let L be the length of is shadow, measured from the man feet to the end of the shadow.
(dL/dt is the rate of the shadow lengthens.
Using similar triangle
6/15=L/(L+x) or
6L+6x=15L
6x=9 L
x=3/2 L
dx/dt= 3/2 dL/dt
well, then dL/dt=2/3 dx/dt= 2/3 5=10/3 ft/sec
Now if you were to ask at what rate the shadow tip was moving , that is dx/dt+dL/dt
let x be the distance from the lamppost to the man (dx/dt=5; given) Let L be the length of is shadow, measured from the man feet to the end of the shadow.
(dL/dt is the rate of the shadow lengthens.
Using similar triangle
6/15=L/(L+x) or
6L+6x=15L
6x=9 L
x=3/2 L
dx/dt= 3/2 dL/dt
well, then dL/dt=2/3 dx/dt= 2/3 5=10/3 ft/sec
Now if you were to ask at what rate the shadow tip was moving , that is dx/dt+dL/dt
Answered by
Heena Patel
let x be the rate of shadow.
x / (x + 5) = 6 / 15
15x = 6(x+5) = 6x + 30
9x = 18
x = 30/9 = 10/3
x / (x + 5) = 6 / 15
15x = 6(x+5) = 6x + 30
9x = 18
x = 30/9 = 10/3
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.