16. Parametric Equations & Polar Coordinates

Calculus with Parametric Curves

16. Parametric Equations & Polar Coordinates

Calculus with Parametric Curves

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Multiple Choice

Find $\frac{d^{3} y}{d x^{3}}$ for the parametric curve at the given point.
$x equals 2 t cubed$ , $y = t^{4}$ , $t equals 1$

$- \frac{4}{9}$

$- \frac{2}{9}$

$\frac{2}{3}$

$- \frac{1}{27}$

Verified step by step guidance

Step 1: Begin by recalling the formula for the third derivative of y with respect to x for a parametric curve. The formula is $ \frac{d^3y}{dx^3} = \frac{d}{dx} \left( \frac{d^2y}{dx^2} \right) $. To compute this, we need $ \frac{dy}{dx} $, $ \frac{d^2y}{dx^2} $, and then differentiate $ \frac{d^2y}{dx^2} $ with respect to x.

Step 2: Compute $ \frac{dy}{dx} $ using the chain rule. For parametric equations $ x = 2t^3 $ and $ y = t^4 $, $ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} $. First, calculate $ \frac{dy}{dt} = 4t^3 $ and $ \frac{dx}{dt} = 6t^2 $. Then, substitute these into $ \frac{dy}{dx} = \frac{4t^3}{6t^2} $. Simplify the expression.

Step 3: Compute $ \frac{d^2y}{dx^2} $. Differentiate $ \frac{dy}{dx} $ with respect to t, and then divide by $ \frac{dx}{dt} $. Use the quotient rule to differentiate $ \frac{dy}{dx} $ with respect to t. The quotient rule states $ \frac{d}{dt} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} $. Apply this to $ \frac{dy}{dx} = \frac{4t^3}{6t^2} $, and then divide the result by $ \frac{dx}{dt} = 6t^2 $.

Step 4: Compute $ \frac{d^3y}{dx^3} $. Differentiate $ \frac{d^2y}{dx^2} $ with respect to t, and then divide by $ \frac{dx}{dt} $. Again, use the quotient rule to differentiate $ \frac{d^2y}{dx^2} $ with respect to t. After obtaining the derivative, divide by $ \frac{dx}{dt} = 6t^2 $. Simplify the resulting expression.

Step 5: Evaluate $ \frac{d^3y}{dx^3} $ at $ t = 1 $. Substitute $ t = 1 $ into the simplified expression for $ \frac{d^3y}{dx^3} $. This will give the value of the third derivative at the given point.