# What is the slope of f(t) = (t^2-tsqrtt,t/2) at t =3?

Jun 15, 2017

$m = - 8.0386$

#### Explanation:

This parametric equation says that the $x$ value of the function $f$ is

$x \left(t\right) = {t}^{2} - t \sqrt{t}$

When $t = 3$

$x \left(3\right) = {3}^{2} - 3 \sqrt{3} = 9 - 3 \sqrt{3} \approx 3.8038$

The derivative is

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

When $t = 3$, the derivative of $x$ becomes

dx/dt=2(3)-(3(1/2(3)^(-1/2)+sqrt(3))~~-0.0622

It also says the $y$ value of the function is

$y \left(t\right) = \frac{t}{2}$

When $t = 2$, $y \left(2\right) = 1$

The derivative of $y$ is

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{1}{2} \approx 0.5$

The slope of the function is

$m = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$ where $\frac{\mathrm{dx}}{\mathrm{dt}} \ne 0$

Plugging in our derivatives from above at $t = 3$, we get

$m = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{0.5}{-} 0.0622 \approx - 8.0386$

We now have ${x}_{1} = 3.8038$, ${y}_{1} = 1$, and $m = - 8.0386$