Two ships (Related Rates)

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jkhaaaan

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Nov 28, 2009
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hi, i have this question for homework and it's not like the ones we've done in class so i'm a little confused. here's the question-

two ships start at the same place but ship A leaves at noon heading east at 12 mph but ship B leaves the same place but at 1 pm heading 120 degrees to the path of ship A. Find the rate of change of the distance between the two ships at 3 pm if ship B has a rate of 5 mph.

so far, i think you have to use the law of cosines since it's not a right triangle but then when i was taking the derivative of the equation, i don't have dtheta/dt and then i had to do the product rule on three terms... is there something i was supposed to plug in before taking the derivative? and do the times matter?
 
jkhaaaan said:
hi, i have this question for homework and it's not like the ones we've done in class so i'm a little confused. here's the question-

two ships start at the same place but ship A leaves at noon heading east at 12 mph but ship B leaves the same place but at 1 pm heading 120 degrees to the path of ship A. Find the rate of change of the distance between the two ships at 3 pm if ship B has a rate of 5 mph.

so far, i think you have to use the law of cosines since it's not a right triangle but then when i was taking the derivative of the equation, i don't have dtheta/dt and then i had to do the product rule on three terms... is there something i was supposed to plug in before taking the derivative? and do the times matter?
Click to expand...

c[sup:1d4zh2rh]2[/sup:1d4zh2rh] = a[sup:1d4zh2rh]2[/sup:1d4zh2rh] + b[sup:1d4zh2rh]2[/sup:1d4zh2rh] - 2ab*cos(C)

The angle C remains constant = 120°

Now continue ....
 
\(\displaystyle At \ 3pm \ ship \ a \ has \ traveled \ 36 \ miles \ and \ ship \ b \ has \ traveled \ 10 \ miles.\)

\(\displaystyle So, \ the \ distance \ between \ them \ is: \ c \ = \ \sqrt(36^{2}+10^{2}-2*10*36*cos(120^{0})) \ = \ about \ 42 \ miles.\)

\(\displaystyle Now, \ we \ are \ given \ that \ \frac{da}{dt} \ = \ 12 \ and \ \frac{db}{dt} \ = \ 5, \ find \ \frac{dc}{dt}.\)

\(\displaystyle c^{2} \ = \ a^{2}+b^{2}-2abcos(120^{0}) \ \implies \ 2c(dc/dt) \ = \ 2a(da/dt)+2b(db/dt)-2cos(120^{0})[a(db/dt)\)

\(\displaystyle +b(da/dt)]\)

\(\displaystyle Now, \ all \ you \ have \ to \ do \ is \ solve \ for \ \frac{dc}{dt}, \ can \ you \ take \ it \ from \ here?\)
 
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