A woman 5 ft tall walks at the rate of 3.5 ft/sec away from a streetlight that is 12 ft above the ground. At what rate is the tip of her shadow moving? At what rate is her shadow lengthening?

Question

anonymous · Answer

i know that 12/5=(x+y)/y but the answer i got after solving and pugging it rate @3.5....i got 42 but the answer is 6 for part a and 6.4 for part b

amistre64 · Answer

we get a lot of streetlight shadow questions lately, i think its a new trend in shadow watching

amistre64 · Answer

we need a relationship that we can work with; similar triangles comes to mind

anonymous · Answer

part b is 2.5

amistre64 · Answer

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mathteacher1729 · Answer

Lots of examples of related rates problems here: http://tutorial.math.lamar.edu/Classes/CalcI/RelatedRates.aspx Example six is probably the most relevant. (It has a nice diagram and explanation)

amistre64 · Answer

$$\frac{12}{5}=\frac{w+s}{s}$$
is a good start yes; all we gotta do is derive it with respect to time to get:
$$0=\frac{s(w'+s')-s'(w+s)}{s^2}$$

its a bit messy but then we solve it for s' for the speed at which the shadow is growing

amistre64 · Answer

$$0=\frac{s(w'+s')-s'(w+s)}{s^2}$$

$$0=s(w'+s')-s'(w+s)$$
$$s'(w+s)=s(w'+s')$$
$$s'=\frac{s(w'+s')}{(w+s)}$$
w'=3.5

amistre64 · Answer

youve got no distance walked to calibrate for the rest of it

amistre64 · Answer

opps, forgot an s' in there dint i

amistre64 · Answer

$$s'(w+s)=s(w'+s')$$
$$s'(w+s)=sw'+ss'$$
$$s'(w+s)-ss'=sw'$$
$$s'((w+s)-s)=sw'$$
$$s'=\frac{sw'}{((w+s)-s)}$$
$$s'=\frac{sw'}{w}$$
that seems better

amistre64 · Answer

so without any further information id way it is:

$$s' = 3.5\frac{s}{w}$$
and the speed of the tip is just:
$$w'+s'$$
$$3.5+3.5\frac{s}{w}$$

amistre64 · Answer

any of this make sense?

anonymous · Answer

kinda

amistre64 · Answer

your missing the part in your question about how far away from the street light she walked

amistre64 · Answer

when you get it; then im sure we can relate the shadow in terms of walking:

$$\frac{12}{5}=\frac{w+s}{s}$$
$$12s=5w+5s$$
$$12s-5s=5w$$
$$s(12-5)=5w$$
$$s=\frac{5w}{7}$$

anonymous · Answer

i get that part

amistre64 · Answer

well, without a distance walked or some other information i got no further i can take this ....