Two ships (Related Rates)

[jkhaaaan]

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?

 

[Deleted member 4993]

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?

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 ....

 

[BigGlenntheHeavy]

$\displaystyle At \ 3pm \ ship \ a \ has \ traveled \ 36 \ miles \ and \ ship \ b \ has \ traveled \ 10 \ miles.$

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

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

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

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

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

 

[jkhaaaan]

thank you both so much! i can take it from there )