1
$\begingroup$

Part 2 Question

The pictures above describes the question. We have to find the rate of change in x-axis direction.

The answer is derived from implicit differentiation and is $4/3$. The process is: [y(t) gives y-axis length (red line) and x(t) gives x-axis length (green line) where t is seconds after initial picture]

$$ \begin{align} &(y(t))^2 + (x(t))^2 &= 5^2 \\ &\implies \frac {d[(y(t))^2 + (x(t))^2]} {dt} &= \frac{d25} {dt} \\ &\implies \frac {2y\frac{dy(t)}{dt} + 2x\frac{dx(t)}{dt}} {dt} &= 0 \\ &\implies {2y\frac{dy(t)}{dt} + 2x\frac{dx(t)}{dt}} &= 0 \\ \end{align} $$

At $t=0$,

$$ \begin{align} &{2(4)(-1) + 2(3)\frac{dx(t)}{dt}} &= 0 \\ &\implies -8+6\frac{dx(t)}{dt} &= 0 \\ &\implies \frac{dx(t)}{dt} &= \frac 86 \\ &\implies \frac{dx(t)}{dt} &= \frac 43 \\ \end{align} $$

But by solving intuitionally: $y(t)$ becomes 0 after $4/1 = 4$ seconds so the $x(t) = 5$ when $t = 4$. Since the rate of change is constant then we can calculate rate of change as change in $x(t)$ from $t=0$ to $t=4$ divided by $4-0=4$

$$ \text{Rate of change = } = \frac {5-3} {4} = \frac 2 4 = \frac 12 $$

This contradicts above result. So which one is correct and why?

One possible answer I thought could be that the rate of change of $x(t)$ is not constant as $y(t)$ is. So is that it?

$\endgroup$
8
  • 1
    $\begingroup$ Why do you think he rate of change of y or x is constant? $\endgroup$
    – Paul
    Commented Dec 8, 2023 at 20:58
  • $\begingroup$ because isn't question tells us that rate of change of y is constant with $-1$ $\endgroup$
    – user1219159
    Commented Dec 8, 2023 at 20:59
  • 1
    $\begingroup$ @PrabhasKumar: but then your equations show that $$ x'=\frac yx,$$ clearly not constant. $\endgroup$ Commented Dec 8, 2023 at 21:00
  • $\begingroup$ @MartinArgerami I also saw that so the rate of change x is infact not constant? $\endgroup$
    – user1219159
    Commented Dec 8, 2023 at 21:01
  • 1
    $\begingroup$ @Paul No thanks. I'll take your word for it. $\endgroup$
    – user1219159
    Commented Dec 8, 2023 at 21:28

0

You must log in to answer this question.