Let $x ( t ) = 1 - e ^ { t } , y ( t ) = e ^ { 2 t }$ be the parametric equations of a curve. Then $\frac { d y } { d x }$ is

A) $e ^ { t }$
B) $e ^ { - t }$
C)-1
D) $e ^ { - 2 t }$
E) $- 2 e ^ { t }$

Question

Let $x ( t ) = 1 - e ^ { t } , y ( t ) = e ^ { 2 t }$ be the parametric equations of a curve. Then $\frac { d y } { d x }$ is

A) $e ^ { t }$
B) $e ^ { - t }$
C)-1
D) $e ^ { - 2 t }$
E) $- 2 e ^ { t }$

Quizplus · Accepted Answer

To find $\frac{dy}{dx}$, we use the chain rule, which gives us $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. First, we find the derivatives of $y(t)$ and $x(t)$ with respect to $t$:- $dx/dt = -e^t$- $dy/dt = 2e^{2t}$Thus, $\frac{dy}{dx} = \frac{2e^{2t}}{-e^t} = -2e^{2t - t} = -2e^t$.

Let x(t)=1−et,y(t)=e2tx ( t ) = 1 - e ^ { t } , y ( t ) = e ^ { 2 t }x(t)=1−et,y(t)=e2t

Let $x ( t ) = 1 - e ^ { t } , y ( t ) = e ^ { 2 t }$