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

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Answer 1 · 2017-06-15T19:37:39

$m=-8.0386$

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

$x(t)=t^2-tsqrt(t)$

When $t=3$

$x(3)=3^2-3sqrt(3)=9-3sqrt(3)~~3.8038$

The derivative is

$dx/dt=2t-(t(1/2t^(-1/2))+sqrt(t))$

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

$y(t)=t/2$

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

The derivative of $y$ is

$dy/dt=1/2~~0.5$

The slope of the function is

$m=(dy)/dx=(dy/dt)/(dx/dt)$ where $dx/dt != 0$

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

$m=(dy)/dx= 0.5/-0.0622~~-8.0386$

We now have $x_1=3.8038$ , $y_1=1$ , and $m=-8.0386$

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