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I was trying to solve the following problem.

A mouse is located at the origin of the coordinate plane and a cat is directly 10 units below it. Then the mouse starts traveling at a constant speed of 6 along the positive x-axis, and the cat starts traveling at a constant speed of 8 but the direction of the cat is always directed towards the mouse at any instant. Find the explicit equation that traces out the path traveled by the cat at each point in time and also find the equation of the path traveled by the cat without the parameter time(i.e. only in terms of x and y).

I tried to do this rather directly as possible. let $\overrightarrow{R}(t)$ be the position of the mouse at each point in time. Let $\overrightarrow{V_c}(t)$ be the velocity of the cat at each point in time and let $\overrightarrow{C}(t)$ be the position of the cat at each point in time. Another piece of information we know is that $||\overrightarrow{V_c}(t)||$ is 8 for all t and $\overrightarrow{C}(0)=-10\hat{j}$.

So I set up the necessary differential equation.$$\frac{d}{dt}\overrightarrow{C}(t)=\frac{8[\overrightarrow{R}(t)-C(t)]}{||\overrightarrow{R}(t)-C(t)||}$$expanding it out you get:$$\frac{d}{dt}[x_c(t)\hat{i}+y_c(t)\hat{j}]=\frac{8[6t-x_c(t)]\hat{i}+8[-y_c(t)]\hat{j}}{\sqrt{(6t-x_c(t))^2+(y_c(t))^2}}$$After this point I am completely clueless what to do how do you solve this differential equation? Is there any known analytic solution to this differential equation or would you solve it numerically Help!!!???

Hammock
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    You set up the differential equation correctly. When you differentiate a vector, the definition is the same as before $$\frac{d}{dt}x_c(t)\hat{i}|{t= t_0} := \lim{h \to 0} \frac{x_c(t_0+h)\hat{i} - x_c(t_0)\hat{i}}{h} = x_c'(t_0)\hat{i}.$$

    For your case, equate the $\hat{i}$ and $\hat{j}$ coeffcients on both sides of the equation. A notation that might help you more is writing $C(t) = (x(t), y(t))$, $C'(t) = (x'(t),y'(t)$ and the right hand side as $8(-6t-x(t), -8y(t))/\sqrt{(-6t-x(t))^2+y(t)^2}$. (so like an ordered pair.) It makes more sense to equate the components.

     – 1mdlrjcmed Aug 02 '23 at 18:19
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    Specifically, you get $x'(t) = \frac{8(6t-x(t))}{\sqrt{(6t-x(t))^2+y(t)^2}}$ and $y'(t) = \frac{8y(t)}{\sqrt{(6t-x(t))^2+y(t)^2}}$ (I think you missed the signs, position of mouse is $(6t, 0)$ and position of cat is $(x(t), y(t))$ so $R(t)-C(t) = (6t-x(t), -y(t))$ – 1mdlrjcmed Aug 02 '23 at 18:24
  • Yeah sorry, I made the edit. The sign changes were my mistake, but I still think in the numerator it should be $-y_c(t)$ because the y component of Vector R is zero and the y component of vector C(t) is y(t) so 0-y(t)=y(t). but I understood how you split it into to components. Now it became more complex, how would find functions x(t) and y(t) to satisfy this differential equation??? Is there any known solution or is it solved numerically. – Hammock Aug 03 '23 at 00:51
  • Yup, you're right about that. The solution is discussed in here https://math.stackexchange.com/questions/244333/cat-dog-problem-using-integration?rq=1 – 1mdlrjcmed Aug 03 '23 at 01:33

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