What is the slope of $f (t) = (t^{2} - t \sqrt{t}, \frac{t}{2})$ $f(t) = (t^2-tsqrtt,t/2)$ at $t = 3$ $t =3$ ?

Question

1789 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-15T19:37:39

$m = - 8.0386$ $m=-8.0386$

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

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

When $t = 3$ $t=3$

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

The derivative is

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

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

$\frac{d x}{d t} = 2 (3) - (3 (\frac{1}{2} {(3)}^{- \frac{1}{2}} + \sqrt{3}) \approx - 0.0622$ $dx/dt=2(3)-(3(1/2(3)^(-1/2)+sqrt(3))~~-0.0622$

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

$y (t) = \frac{t}{2}$ $y(t)=t/2$

When $t = 2$ $t=2$ , $y (2) = 1$ $y(2)=1$

The derivative of $y$ $y$ is

$\frac{d y}{d t} = \frac{1}{2} \approx 0.5$ $dy/dt=1/2~~0.5$

The slope of the function is

$m = \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $m=(dy)/dx=(dy/dt)/(dx/dt)$ where $\frac{d x}{d t} \neq 0$ $dx/dt != 0$

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

$m = \frac{d y}{d x} = \frac{0.5}{- 0.0622} \approx - 8.0386$ $m=(dy)/dx= 0.5/-0.0622~~-8.0386$

We now have $x_{1} = 3.8038$ $x_1=3.8038$ , $y_{1} = 1$ $y_1=1$ , and $m = - 8.0386$ $m=-8.0386$

What is the slope of f(t) = (t^2-tsqrtt,t/2)f(t)=(t2−t√t,t2)f(t) = (t^2-tsqrtt,t/2) at t =3t=3t =3?