Module 25 - Parametric Equations | ||||||||||
Introduction | Lesson 1 | Lesson 2 | Lesson 3 | Self-Test | ||||||||||
Lesson 25.2: Chain Rule for Parametric Equations | ||||||||||
This lesson investigates the procedure to find derivatives, such as
The Chain Rule Suppose a curve is defined by the parametric equations x = f(t) y = g(t)
The Chain Rule states that the derivative
![]() Finding First Derivatives The values of the derivatives dy/dt, dx/dt, and dy/dx for a set of parametric equations can be found using the TI-83. Suppose you wish to find the values of the derivatives when t = 0.5 for the parametric equations y(t) = tan t
Recall that
Finding Derivatives from a Graph The derivatives at a point on the graph of a parametric curve can be found by using the derivative features of the CALC menu on the Graph screen.
![]() The graph is a hyperbola. The two diagonal lines are not part of the graph. The two lines are similar to the vertical lines that often appear in graphs of functions that have vertical asymptotes.
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The calculator returns to the graph.
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![]() The value of dy/dx at t = 0.5 is approximately 2.0858. That is, the y-values are increasing about 2.0858 times as fast as the x-values. 25.2.1 Use the CALC menu to find the value of dy/dt and dx/dt at t = 0.5. Do these values satisfy the Chain Rule for parametric equations? Click here for the answer. Finding the Equation of the Tangent Line
The slope of the tangent line is dy/dx
![]() The coordinates of the point on the curve when t = 0.5 are approximately (1.1395, 0.5463). Using the point just found and the value of dy/dx for the slope of the line, the equation of the tangent line is Graph this function parametrically.
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![]() 25.2.2 Graph the curve given by the parametric equations below and find the equation of the tangent line to the curve when t = 2. y = 1 - cos(t) Click here for the answer. |
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