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?

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

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

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

part b is 2.5

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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)

\[\frac{12}{5}=\frac{w+s}{s}\] 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

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

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

opps, forgot an s' in there dint i

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

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

any of this make sense?

kinda

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

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

i get that part

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

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