How do you differentiate the following parametric equation: $x (t) = - e^{t} - 4 t, y (t) = - 5 t^{2} + 2$ $x(t)=-e^t-4t, y(t)= -5t^2+2$ ?

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Answer 1 · 2017-09-03T12:01:35

$\frac{d y}{d x} = \frac{10 t}{e^{t} + 4}$ $dy/dx = ( 10t ) / ( e^t+4 )$

We have:

$x (t) = - e^{t} - 4 t$ $x(t) = -e^t-4t$
$y (t) = - 5 t^{2} + 2$ $y(t) = -5t^2+2$

Differentiating each equation wrt $t$ $t$ we get:

$x'(t) = -e^t-4$
$y'(t) = -10t$

By the chain rule rule we have:

$dy/dx = (dy/dt) / ( dx/dt)$

$\ \ \ \ \ \= ( y'(t) ) / ( x'(t) )$

$\ \ \ \ \ \= ( -10t ) / ( -e^t-4 )$

$\ \ \ \ \ \= ( 10t ) / ( e^t+4 )$

How do you differentiate the following parametric equation: x(t)=-e^t-4t, y(t)= -5t^2+2 x(t)=−et−4t,y(t)=−5t2+2 x(t)=-e^t-4t, y(t)= -5t^2+2 ?