Question

Could someone help me with a Calc1 Related Rates problem, please?

Answer 1

The picture was a condensed from my notes, but I don't know how to do it.
the actual problem was along the lines of:
Bobby Teenager is walking away from a lamp post... the rate at which his shadow is changing is 1/3 the rate he is walking... the lamp post is 20ft high...
a) Explain how Bobby Teenager is 5ft tall.
b) Find the rate at which (theta) is changing when Bobby Teenager is 12ft away from the lamp post. and he is walking at a speed of 5 ft/min.

Answer 2

strategy, find a relationship between the parts ...

take the implicit derivative of the relationship

fill in what you know to find what you do not know

Answer 3

similar triangles seems to be an appropriate relationship

Answer 4

umm, I had interpreted the question as I would have to do something to find that his height is 5... ?

Answer 5

you said you thought the actual problem went like such and such .. which may differ from how the actual problem is. We can try to see if finding h=5 is applicable

Answer 6

if we make variables for all of it, then fill in the specifics we get something like: l for lamp, h for height, s for shadow, w for distance walked...

\[\frac{l}{w+s}=\frac{h}{s}\]

Answer 7

seems like all we know from the start is l and s' lets keep that in mind and just jump in the derivative

\[\frac{l}{w+s}=\frac{h}{s}\]
\[ls=h(w+s)\]
\[l's+ls'=h'(w+s)+h(w'+s')\]
since l and h are constant, there derivatives are 0
\[ls'=h(w'+s')\]
filling in what we know
\[20/3=h(w'+1/3)\]
it seems we need a way to express w' in terms of what we know

Answer 8

helps if you read it right :) s'=w'/3

Answer 9

\[ls'=h(w'+s')\]
\[20w'/3=h(w'+w'/3)\]
\[20w'/3=h(4w'/3)\]
\[20=h(4)\]
yeah, h=20/4 = 5

Answer 10

for the next one we can use a trig relationship, and work the same strategy

Answer 11

okay, I see it now! thankyou :)

Answer 12

youre welcome, of course instead of w+s we can use the d for the distance,
and d' = w'+s' = 4w'/3

Answer 13

so (theta)' = (-l (4w'/3))/((sec(theta))^2)(d^2) ?

Answer 14

so theta' = (25 (cos(theta))^2) / 48 ?
how would find cos(theta)?

Answer 15

I got -1025/3 or -314.666... for part b. would that be correct?
@amistre64

Answer 16

part b eh ....

tan(a) = l/d

sec^2(a) a' = (dl'-d'l)/d^2

l is constant so l' = 0

sec^2(a) a' = -d'l/d^2

d' = w'+s' = 20/3
l = 20

d and a can be similarly calculated

20 5
--- = -----
d d-12

15d = 240 ; d = 16

a = arctan(20/16)

putting it all together we get:

a' = -20(20)
-------- cos^2(arctan(5/4))
3

im getting about -52 ...

Answer 17

what happened to the d^2 in the denominator?

sorry it took me a while to come back, my internet died...

Answer 18

for future reference: indeed the d^2 in the denominator had been forgotten
therefore correct answer was -> -0.205 radians/min
(for the second part...)

and thank you once again amistre64! :D