| Module 25 - Parametric Equations | ||||||||||
| Introduction | Lesson 1 | Lesson 2 | Lesson 3 | Self-Test | ||||||||||
| Lesson 25.2: Chain Rule for Parametric Equations | ||||||||||
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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.
The calculator returns to the graph.
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.
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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