Question

Tangent Lines to Parametrized CurvesIn Exercises 1−14, find an equation for the line tangent to the curve at the point defined by the given value of t. Also, find the value of d²y/dx² at this point.x = t + eᵗ, y = 1 − eᵗ, t = 0

Accepted Answer

First, find the coordinates of the point on the curve at $$ t = 0  by $$substituting $$ t = 0 $$ into the parametric equations $$ x = t + e^{t} $$ and $$ y = 1 - e^{t} . $$This gives the point $$ (x_0, y_0) . $$Next, compute the derivatives $$ \frac{dx}{dt} $$ and $$ \frac{dy}{dt}  by $$differentiating each parametric equation with respect to $$ t . $$Then, find the slope of the tangent line $$ \frac{dy}{dx}  at$$ $$ t = 0 $$ using the chain rule for parametric curves: $$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} . $$Use the point-slope form of a line to write the equation of the tangent line at the point $$ (x_0, y_0) $$ with slope $$ \frac{dy}{dx} $$: $$ y - y_0 = \left( \frac{dy}{dx} \right)(x - x_0) . To $$find the second derivative $$ \frac{d^2 y}{dx^2}  at$$ $$ t = 0 $$, first compute $$ \frac{d}{dt} \left( \frac{dy}{dx} \right) $$, then divide by $$ \frac{dx}{dt} $$ using the formula: $$ \frac{d^2 y}{dx^2} = \frac{\frac{d}{dt} \left( \frac{dy}{dx} \right)}{\frac{dx}{dt}} . $$Evaluate this expression at $$ t = 0 .$$

Tangent Lines to Parametrized Curves

In Exercises 1−14, find an equation for the line tangent to the curve at the point defined by the given value of t. Also, find the value of d²y/dx² at this point.

x = t + eᵗ, y = 1 − eᵗ, t = 0

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Key Concepts

Parametric Equations and Curves

Finding the Tangent Line to a Parametric Curve

Second Derivative d²y/dx² for Parametric Curves