What is the slope of $f(t) = (t-2t+4,t^2)$ at $t =0$ ?

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Answer 1 · 2016-11-22T00:28:51

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Define $x(t), y(t)$ as follows

${ (x(t)=t-2t+4=4-t), (y(t)=t^2) :} => f(t) = (x(t), y(t))$

Differentiating $x(t)$ wrt $t$ we get:

$dot x(t) = dx/dt = -1$

Differentiating $y(t)$ wrt $t$ we get:

$dot y(t) = dy/dt=2t$

Then the applying the chain rule;

$f'(t) = dy/dx = dy/dt*dt/dx = (dy/dt)/(dx/dt)$
$:. f'(t) = ( 2t ) / ( -1 )$
$:. f'(t) = -2t$

So when $t=0$ we get:

$f'(0) = 0$

Hence, The slope of $f(t)$ is $0$ at $t=0$

What is the slope of f(t) = (t-2t+4,t^2)f(t) = (t-2t+4,t^2) at t =0t =0?