Boat B Must Head 56.4° South of West

[markosheehan]

a boat A is heading south at 30 km/hr. its spotted a boat B which is 18 km due east of boat A. boat B has a max speed of 25 km/hr and wishes to get a close as possible to boat A. what direction should boat B head in.

so i initially let boat A be the origin heading south. boat B starts at (18,0) and heads south of west. so i let 18-25sina=0 solving this gives 46 degrees. the answer is 56.4 south of west. i can explain what i did in more detail if anyone needs it.

 

[MarkFL]

I think what I would do is orient a coordinate system such that south is up and west is to the right. At time $t=0$, have boat A at the origin, and boat B at $(-18,0)$. Then, parametrize the motion of the two boats:

Boat A:

\(\displaystyle x_A(t)=0\)

\(\displaystyle y_A(t)=30t\)

Boat B:

\(\displaystyle x_B(t)=25\cos(\theta)t-18\)

\(\displaystyle y_B(t)=25\sin(\theta)t\)

Now, the square of the distance $D$ between the boats is:

\(\displaystyle D^2=f(\theta,t)=\left(25\cos(\theta)t-18\right)^2+\left(30t-25\sin(\theta)t\right)^2\)

So, we want to find any critical points:

\(\displaystyle f_{\theta}(\theta,t)=2\left(25\cos(\theta)t-18\right)(-25\sin(\theta)t)+2\left(30t-25\sin(\theta)t\right)(-25\cos(\theta)t)=0\)

This reduces to (discarding the boundary $t=0$):

\(\displaystyle t=\frac{3}{5}\tan(\theta)\)

\(\displaystyle f_{t}(\theta,t)=2\left(25\cos(\theta)t-18\right)(25\cos(\theta))+2\left(30t-25\sin(\theta)t\right)(30-25\sin(\theta))=0\)

This reduces to:

\(\displaystyle 61t-18\cos(\theta)-60\sin(\theta)t=0\)

Substituting for $t$, we obtain:

\(\displaystyle 61\left(\frac{3}{5}\tan(\theta)\right)-18\cos(\theta)-60\sin(\theta)\left(\frac{3}{5}\tan(\theta)\right)=0\)

\(\displaystyle 183\sin(\theta)-90\cos^2(\theta)-180\sin^2(\theta)=0\)

\(\displaystyle 90\sin^2(\theta)-183\sin(\theta)+90=0\)

\(\displaystyle 3\left(5\sin(\theta)-6\right)\left(6\sin(\theta)-5\right)=0\)

Discarding the root outside of the range of the sine function, we are left with:

\(\displaystyle \sin(\theta)=\frac{5}{6}\implies\theta\approx56.44269023807928^{\circ}\)

 

[markosheehan]

I think what I would do is orient a coordinate system such that south is up and west is to the right. At time $t=0$, have boat A at the origin, and boat B at $(-18,0)$. Then, parametrize the motion of the two boats:

Boat A:

\(\displaystyle x_A(t)=0\)

\(\displaystyle y_A(t)=30t\)

Boat B:

\(\displaystyle x_B(t)=25\cos(\theta)t-18\)

\(\displaystyle y_B(t)=25\sin(\theta)t\)

Now, the square of the distance $D$ between the boats is:

\(\displaystyle D^2=f(\theta,t)=\left(25\cos(\theta)t-18\right)^2+\left(30t-25\sin(\theta)t\right)^2\)

So, we want to find any critical points:

\(\displaystyle f_{\theta}(\theta,t)=2\left(25\cos(\theta)t-18\right)(-25\sin(\theta)t)+2\left(30t-25\sin(\theta)t\right)(-25\cos(\theta)t)=0\)

This reduces to (discarding the boundary $t=0$):

\(\displaystyle t=\frac{3}{5}\tan(\theta)\)

\(\displaystyle f_{t}(\theta,t)=2\left(25\cos(\theta)t-18\right)(25\cos(\theta))+2\left(30t-25\sin(\theta)t\right)(30-25\sin(\theta))=0\)

This reduces to:

\(\displaystyle 61t-18\cos(\theta)-60\sin(\theta)t=0\)

Substituting for $t$, we obtain:

\(\displaystyle 61\left(\frac{3}{5}\tan(\theta)\right)-18\cos(\theta)-60\sin(\theta)\left(\frac{3}{5}\tan(\theta)\right)=0\)

\(\displaystyle 183\sin(\theta)-90\cos^2(\theta)-180\sin^2(\theta)=0\)

\(\displaystyle 90\sin^2(\theta)-183\sin(\theta)+90=0\)

\(\displaystyle 3\left(5\sin(\theta)-6\right)\left(6\sin(\theta)-5\right)=0\)

Discarding the root outside of the range of the sine function, we are left with:

\(\displaystyle \sin(\theta)=\frac{5}{6}\implies\theta\approx56.44269023807928^{\circ}\)

thanks for your reply. just wondering is there a way using triangles to solve this using the sine rule do you know?

 

[MarkFL]

Being that it is an optimization problem, on a function in two variables, the way I worked it using the calculus is the most straightforward way I know of. :D

In what course was this problem given?

 

[markosheehan]

It's in a relative velocity chapter in math physics

 

[MarkFL]

It's in a relative velocity chapter in math physics

Is this a calculus-based physics course?

 

[markosheehan]

theres no calculus in the course. I am a bit confused on your answer. so i get that you are squaring the distance in the y direction between the boats and you are squaring the distance in the x direction between the boats and you are letting it equal to zero because the the boat B is aiming to intercept boat A.

however i don't understand your method of factorising (25cos(a)t-18)^2 + (30t-25sin(a)t)^2 =0

also i was thinking an easier way to solve it would be to say the distance traveled by the boats in the y direction is 25sinat=30t so 25sina=30 sina=30/25 sadly this doesn't work. why doesn't it work do you know?