1. ## interception

a cruiser is 12 km due west of a destroyer . the cruiser is travelling in a direction 60 degrees N of E at 30 km hr^-1 and the destroyer is travelling at 15 km hr^-1. find the direction the destroyer must travel in order to get as close as possible to the cruiser

2. ## Re: interception

Let's use vectors. The cruiser's initial position is $0\hat{i}+0\hat{j}$. The destroyer's original position is $12\hat{i}+0\hat{j}$.

The cruiser's position after $t$ hours will be:

$30t\cos 60^\circ \hat{i} + 30t\sin 60^\circ \hat{j}$

The destroyer's position after $t$ hours will be:

$12+15t\cos \theta \hat{i} + 15t\sin \theta \hat{j}$

The distance between them will be:

$\sqrt{(12+15t\cos \theta - 15t)^2+(15t\sin \theta - 15t\sqrt{3})^2}$

You are minimizing with respect to two variables.

3. ## Re: interception

I believe that the destroyer should travel at 60 degrees North of West. I found that by taking the partial derivative with respect to $t$, setting it equal to zero, solving for $t$, plugging it back into the distance formula, then since I had a function of one variable, I minimized with respect to $\theta$ and got $\theta = \dfrac{2\pi}{3}$, which is $120^\circ$, or 60 degrees North of West. However, this is not a guaranteed minimum. This may not even be a local minimum. There are certain conditions that must be met in order for this to be a minimum. I get that $t=24\text{ mins}$ gives the minimum by this method. That would be 6km apart after 24 minutes. I am not sure if you can get them any closer together, but I certainly did not prove that this was, indeed, the minimum.

4. ## Re: interception

that is the right answer but im not sure how you get it. when i multiply out the expression you gave for the distance between them i get
√{ 225t^2(cosa)+t^2*(225cos^2 a)+225t^2+(360cosa)*t -360t+144 +225t^2(sin^2 a)-450√3*t^2(sina)+675t^2 }
I am not sure how to differentiate this as there is 2 variables

thanks

5. ## Re: interception

I treated $\theta$ as a constant and differentiated WRT $t$.

6. ## Re: interception

Did I multiply it out correctly? Should I use the chain rule and not multiply it out.?
I still not working out for me

7. ## Re: interception

I used Wolframalpha to find derivatives. The way I did it makes it a very nontrivial problem. Perhaps there is a geometric interpretation that simplifies it, but it did not jump out at me.

8. ## Re: interception

does any one else know how to do this problem?

9. ## Re: interception

Do not multiply out. Just use the chain rule.

$D_t = \dfrac{1}{2}\left(2(12+15t \cos \theta - 15t)(15\cos \theta - 15)+2(15t\sin \theta - 15t\sqrt{3})(15\sin \theta - 15\sqrt{3})\right)\left((12+15t\cos \theta - 15t)^2+(15t\sin \theta - 15t\sqrt{3})^2\right)^{-1/2} = 0$

$(12+15t\cos \theta - 15t)(15\cos \theta - 15) + (15t\sin \theta - 15t\sqrt{3})(15\sin \theta-15\sqrt{3}) = 0$

$(12+15t\cos \theta - 15t)(15\cos \theta - 15)+t(15\sin \theta - 15\sqrt{3})^2 = 0$

$12(15\cos \theta-15)+t(15\cos \theta-15)^2+t(15\sin \theta-15\sqrt{3})^2 = 0$

$t = \dfrac{12(15-15\cos \theta)}{(15\cos \theta-15)^2+(15\sin \theta-15\sqrt{3})^2} = \dfrac{4}{5}\dfrac{1-\cos \theta}{(\cos \theta-1)^2+(\sin \theta-\sqrt{3})^2}$

Plug that in for $t$ in the original equation for the distance and then minimize with respect to $\theta$.

10. ## Re: interception

You get

$12\sqrt{\left(1-\dfrac{\sin^2 \theta}{(1-\cos \theta)^2+(\sqrt{3}-\sin \theta)^2}\right)^2+\left(\dfrac{(\sin \theta-\sqrt{3})(1-\cos \theta)}{(1-\cos \theta)^2+(\sqrt{3}-\sin \theta)^2}\right)^2}$

Differentiate that with respect to $\theta$ and set it equal to zero. You should get $\theta = \dfrac{2\pi}{3}$.

11. ## Re: interception

Does not work with partial derivatives...

12. ## Re: interception

thanks for doing this work but my knowledge of calculus is not that advanced

13. ## Re: interception

Then there is probably something with circles and triangles that will yield the same result. I yield the thread to those more knowledgeable in such things.

14. ## Re: interception

Had to recall plotting relative motion on a maneuvering board from my Navy days ...

Cruiser is at the origin, relative motion of the destroyer is the vector $\color{red}{\Delta r}$ in the attached diagram.

15. ## Re: interception

ive tired to solve it another way. i should of mentioned its a question from a relative velocity chapter.

Vc =15sinai+15sinaj

Vd=15i+15√3j

Vcd =(15cosa-15)i+(15sina-15√3)
if you look at skeeter diagram this is this represents the line labelled △r

in my diagram we want x to be a minimum so tan theta should be a minimum

to minimise tana : d(tana)/d theta =0

so i use the chain rule

tan a = (15sina-15√3)/(15cosa-15)

so i simply this and in the chain rule let u=3sina-3√3 and v=3cosa-3

so eventually i get cosa+√3sina=1

i am doing something wrong but i know this method does work as i have seen it work on similar questions.
where am i going wrong or could someone do out the question in the way i am trying to do it correctly?

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