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Ch.12 - Parametric and Polar Curves
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh.12 - Parametric and Polar CurvesProblema 12.1.69a
Capitolo 12, Problema 12.1.69a

67–72. Derivatives Consider the following parametric curves.
a. Determine dy/dx in terms of t and evaluate it at the given value of t.


x = cos t, y = 8 sin t; t = π/2

Guida verificata passo dopo passo
1
Recall that for parametric curves defined by \(x = x(t)\) and \(y = y(t)\), the derivative \(\frac{dy}{dx}\) can be found using the chain rule as \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
Given the parametric equations \(x = \cos t\) and \(y = 8 \sin t\), compute the derivatives with respect to \(t\): \(\frac{dx}{dt} = -\sin t\) and \(\frac{dy}{dt} = 8 \cos t\).
Substitute these derivatives into the formula for \(\frac{dy}{dx}\) to get \(\frac{dy}{dx} = \frac{8 \cos t}{-\sin t} = -8 \frac{\cos t}{\sin t}\).
Evaluate \(\frac{dy}{dx}\) at the given value \(t = \frac{\pi}{2}\) by substituting \(t\) into the expression \(-8 \frac{\cos t}{\sin t}\).
Simplify the expression after substitution to find the slope of the curve at \(t = \frac{\pi}{2}\).

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Parametric Equations

Parametric equations express the coordinates of points on a curve as functions of a parameter, often denoted as t. Instead of y as a function of x, both x and y depend on t, allowing the description of more complex curves. Understanding how to work with these equations is essential for analyzing motion and curves in calculus.
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Parameterizing Equations

Derivative of Parametric Curves (dy/dx)

For parametric curves, the derivative dy/dx is found by dividing the derivative of y with respect to t by the derivative of x with respect to t, i.e., (dy/dt) / (dx/dt). This method allows us to find the slope of the tangent line to the curve at a specific parameter value, even when y is not explicitly given as a function of x.
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Differentiation of Parametric Curves

Evaluating Derivatives at Specific Parameter Values

After finding the general expression for dy/dx in terms of t, evaluating it at a specific value of t gives the slope of the tangent line at that point on the curve. This step involves substituting the given t-value into the derivative expressions and simplifying to find the numerical slope.
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