100 followers special

Yes I agree with Ronak.There is a similar question in IE Irodov.

Figure

$dx/dt$ in the rate at which separation between dog and rabbit decreases. From the figure in vertical direction

$-dx=(2v-vcos\theta )dt$

$\int _{ 300 }^{ 0 }{ -dx } =\int _{ 0 }^{ t }{ (2v-vcos\theta )dt }$

$300=2vt-v\int _{ 0 }^{ t }{ (cos\theta )dt }$ ..........(1)

in horizontal direction

$vt=2v\int _{ 0 }^{ t }{ (cos\theta )dt }$

$\int _{ 0 }^{ t }{ (cos\theta )dt } =\frac { t }{ 2 }$

putting this value in eq (1)

$300=2vt-\frac { vt }{ 2 }$

$t=\frac { 200 }{ v }$

On multiplying this velocity of dog we will get the distance= 400

First of all, I'd like to thank all the 100 followers that have seen me as worthy of being followed. I have more awesome problems coming up in the future, so STAY TUNED XD

And now, a tribute:

ss

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Hope that's all the time for you.

So, back to our problem. I'd like to see if this problem is fool-proof (no shortcuts or weird tricks). This is actually the 4th part of the question, and thus having skipped the first 3 that guide to answer the 4th, making it more difficult. But it's not impossible to answer this question without answering the first 3.

I've taken this problem from the Differential Equations chapter of Stewart's Calculus 6E. It will be posted somewhere on a wall in my school for a 50$ prize. Obviously this is directed towards the math teachers (for reasons), whom I wish to embarrass. I know cannot handle anything of this difficulty. But I can (with the solutions manual at my disposal xD).

So please check this out for me and tell me if it is a difficult one. (I didn't exactly dive into it, just skimmed). The solution sure seems not straight-forward. Here are the 3 sub-problems for those who are wondering:

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The posted problem (here's the exact wording: Suppose that the dog in Problem 9 runs twice as fast as the rabbit. Find a differential equation for the path of the dog. Then solve it to find the point where the dog catches the rabbit. ) is a modified continuation of the original problem, where in the original the dog is as fast as the rabbit.

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(a) Show that the dog's path is the graph of the function $y=f(x)$ , where $y$ satisfies the differential equation

$x\frac{d^2x}{dx^2}=\sqrt{1+(\frac{dy}{dx})^2}$

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(b) Determine the solution of the equation in part (a) that satisfies the initial conditions $y=y'=0$ when $x=L$ . [ Hint : Let $z=dy/dx$ in the differential equation and solve the resulting first-order equation to find $z$ ; then integrate $z$ to find $y$ .]

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(c) Does the dog ever catch the rabbit?

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I shall post solutions upon request.

Cheers