Mathematics OpenStudy (anonymous):

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? OpenStudy (anonymous):

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 OpenStudy (amistre64):

we get a lot of streetlight shadow questions lately, i think its a new trend in shadow watching OpenStudy (amistre64):

we need a relationship that we can work with; similar triangles comes to mind OpenStudy (anonymous):

part b is 2.5 OpenStudy (amistre64):

|dw:1319667372071:dw| OpenStudy (mathteacher1729):

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) OpenStudy (amistre64):

$\frac{12}{5}=\frac{w+s}{s}$ 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 OpenStudy (amistre64):

$0=\frac{s(w'+s')-s'(w+s)}{s^2}$ $0=s(w'+s')-s'(w+s)$ $s'(w+s)=s(w'+s')$ $s'=\frac{s(w'+s')}{(w+s)}$ w'=3.5 OpenStudy (amistre64):

youve got no distance walked to calibrate for the rest of it OpenStudy (amistre64):

opps, forgot an s' in there dint i OpenStudy (amistre64):

$s'(w+s)=s(w'+s')$ $s'(w+s)=sw'+ss'$ $s'(w+s)-ss'=sw'$ $s'((w+s)-s)=sw'$ $s'=\frac{sw'}{((w+s)-s)}$ $s'=\frac{sw'}{w}$ that seems better OpenStudy (amistre64):

so without any further information id way it is: $s' = 3.5\frac{s}{w}$ and the speed of the tip is just: $w'+s'$ $3.5+3.5\frac{s}{w}$ OpenStudy (amistre64):

any of this make sense? OpenStudy (anonymous):

kinda OpenStudy (amistre64):

your missing the part in your question about how far away from the street light she walked OpenStudy (amistre64):

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

i get that part OpenStudy (amistre64):

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

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