# What is the instantaneous velocity of an object moving in accordance to  f(t)= (sqrt(t+2),t+4)  at  t=1 ?

Feb 18, 2017

$\sqrt{2}$

#### Explanation:

Take the derivative of $\sqrt{t + 2}$. You should get:

$\frac{1}{2 \sqrt{t + 2}}$

Plug in 1 to get $\frac{1}{2} \sqrt{2}$.

Now take the derivative of $t + 4$. You should get 1 so plugging in 1 will get you 1.

Now that you know that it has a velocity of $\frac{1}{2} \sqrt{2}$. in the x direction and 1 in the y direction, you need to find the resulting velocity from that. Use Pythagorean theorem:

${a}^{2} + {b}^{2} = {c}^{2}$

${\left(\frac{1}{2 \sqrt{2}}\right)}^{2} + {1}^{2} = {c}^{2}$

$\frac{1}{4 \cdot 2} + 1 = {c}^{2}$

$c = \sqrt{\frac{9}{8}}$

Feb 18, 2017

Instantaneous velocity is given by the vector:

$\vec{v} = \left\langle\frac{\sqrt{3}}{6} , 1\right\rangle$, or, $\frac{\sqrt{3}}{6} \hat{i} + \hat{j}$, or, $\left(\begin{matrix}\frac{\sqrt{3}}{6} \\ 1\end{matrix}\right)$

#### Explanation:

We have:

$f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$ where $x \left(t\right) = \sqrt{t + 1}$, $y \left(t\right) = t + 4$

Then;

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{1}{2} {\left(t + 2\right)}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{t + 2}}$
$\frac{\mathrm{dy}}{\mathrm{dt}} = 1$

So, when $t = 1$:

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{1}{2 \sqrt{1 + 2}} = \frac{1}{2 \sqrt{3}} = \frac{\sqrt{3}}{6}$
$\frac{\mathrm{dy}}{\mathrm{dt}} = 1$

And so the instantaneous velocity is given by the vector:

$\vec{v} = \left\langle\frac{\sqrt{3}}{6} , 1\right\rangle$, or, $\frac{\sqrt{3}}{6} \hat{i} + \hat{j}$, or, $\left(\begin{matrix}\frac{\sqrt{3}}{6} \\ 1\end{matrix}\right)$

If we want the instantaneous speed , it is given by:

$v = | | \vec{v} | |$
$\setminus \setminus \setminus = \sqrt{\frac{3}{36} + 1}$
$\setminus \setminus \setminus = \sqrt{\frac{13}{12}}$
$\setminus \setminus \setminus \approx 1.04083$