Related Rates of two ships

1. Jan 7, 2005

Hello all

I need help with the following question

Two ships A and B are sailing away from the point O along routes such that angle AOB = 120 degrees. How fast is the distance between them changing if, at a certain instant OA = 8 mi. OB = 6 ,i., ship A is sailing at the rate of 20 mi/hr, and ship B at the rate of 30 mi/hr.

I drew an obtuse triangle with the angle 120 degrees. Then I tried using the law of cosines to find the sides. I want to find dY / dt.

Any help is greatly appreciated

Thanks

2. Jan 7, 2005

dextercioby

HINT:Express the area of the triangle in two distinct ways.Use the gact that the angle AOB is constant.

Daniel.

3. Jan 7, 2005

Diane_

Let the sides of the triangle be a, b and c, with c being the straight-line distance between the two ships. Note, then, that you have a, b, da/dt and db/dt. You want dc/dt. All you need is a relationship between a, b, and c - then take the time-derivative of that relationship and you'll be there, minus a little algebra. As you've indicated, you already know how to get that relationship.

4. Jan 7, 2005

HallsofIvy

Staff Emeritus
dextercioby, could you give more detail? Using area sounds interesting but I don't see how it works.

Yes, courtrigrad, using the cosine law is the way I, at least, would do the problem. As Diane suggested, call the length of OA, a, the length of OB, b, and the length of AB, Y (since you used that label). The cosine law says that Y2= a2+ b2- 2abcos(120)= a2+ b2- &radic;(3)ab. To convert that "static" equation into an equation for dY/dt, differentiate both sides (you will need to use "implicit" differentiation). Evaluate with a= 8, b= 6, da/dt= 20, db/dt= 30. You can use the cosine law to find the correct value of Y to put into the equation.

Last edited: Jan 7, 2005
5. Jan 7, 2005

dextercioby

$$S(t)=OB(t)\cdot OA(t) cdot \sin 120=AB(t)\cdot h_{tr.}(t)$$ (1)
There was a '2',but got simplified.Take time derivative
$$[\frac{dOB(t)}{dt}OA(t)+OB\frac{dOA(t)}{dt}]\sin 120=\frac{dAB(t)}{dt}h_{tr.}(t)+AB(t)\frac{dh_{tr.}(t)}{dt}$$(1)

Unfortunately the triagle is not isosceles,so the numbers won't be very pretty.With a little bit of trigonometry and geometry,one finds
$${dh_{tr.}(t)}{dt}=\frac{3\sqrt{39}}{13}\frac{dOA(t)}{dt}$$ (2)
$$AB(t)=2\sqrt{13}$$ (3)
$$h(t)=\frac{24\sqrt{39}}{13}$$ (4)

Eq.(1) pp.(4) give:
$$\frac{dOA(t)}{dt}=\frac{1}{h}\{\frac{\sqrt{3}}{2}[OA(t)\frac{dOB(t)}{dt}+OB(t)\frac{dOA(t)}{dt}]-AB\frac{dh(t)}{dt}\}=\frac{5\sqrt{13}}{2} mph$$ (5)

Daniel.

PS.Maybe your method is better.Mine seemed more intuitive.TO ME.

Last edited: Jan 7, 2005
6. Jan 7, 2005

BobG

You mean ".... = a2+ b2+ab", don't you? (I think you accidentally slipped in the sine of 120 instead of the cosine).

7. Jan 7, 2005

I think Halls is right

c^2 = a^2 + b^2 - 2ab cos C

8. Jan 7, 2005

BobG

Yes, this part is right. The angle stays constant (don't make the mistake of taking the derivative of cos C). Since it's constant, the equation could be simplified to c^2=(a^2 + b^2 +ab) prior to differentiating.

9. Jan 8, 2005

HallsofIvy

Staff Emeritus
Oops! Yes, the cos(120)= -1/2, not &radic;(3)/2. The formula should be
c2= a2+ b2+ ab as BobG said.