Find $x(t)$ given $\frac{dx}{dt}$ and $\frac{dy}{dt}$

Question

For $0 \leq t \leq 1$ , a particle is moving along a curve so that its position at time $t$ is $(x(t),y(t))$. At time $t=0$ , the particle is at position $(0,0)$ .

We are given that

$$\frac{dx}{dt} = \frac{t}{\sqrt {1+t^2}}$$

$$\frac{dy}{dt} = \sqrt {\frac{1-t^2}{1+t^2}}$$

Find $x(t)$

Here is my work:

$$x(t)=\int \frac{dx}{dt}= \int \frac{t}{\sqrt{1+t^2}}dt$$

Substituting:

$$u=t^2+1 \implies du= 2t dt$$

We get;

$$x(t) = \int \frac{t}{\sqrt{1+t^2}}dt = \sqrt{t^2+1}+C$$

Since it is asking for $x(t)$ and not $y(t)$ I believe I'm ok

Answer 1

Your answer is alright except for the $+ C$, notice that the question gave you that $x(0)=y(0)=0$, so that gives $$x(t)=\sqrt{t^2+1}-1$$