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Chap10_Sec1

# Chap10_Sec1 - 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES...

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PARAMETRIC EQUATIONS PARAMETRIC EQUATIONS AND POLAR COORDINATES AND POLAR COORDINATES 10

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PARAMETRIC EQUATIONS & POLAR COORDINATES We have seen how to represent curves by parametric equations. Now, we apply the methods of calculus to these parametric curves.
10.2 Calculus with Parametric Curves In this section, we will: Use parametric curves to obtain formulas for tangents, areas, arc lengths, and surface areas. PARAMETRIC EQUATIONS & POLAR COORDINATES

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TANGENTS In Section 10.1, we saw that some curves defined by parametric equations x = f ( t ) and y = g ( t ) can also be expressed—by eliminating the parameter—in the form y = F ( x ). See Exercise 67 for general conditions under which this is possible.
If we substitute x = f ( t ) and y = g ( t ) in the equation y = F ( x ), we get: g ( t ) = F ( f ( t )) So, if g , F , and f are differentiable, the Chain Rule gives: g’ ( t ) = F’ ( f ( t )) f’ ( t ) = F’ ( x ) f’ ( t ) TANGENTS

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If f’ ( t ) ≠ 0, we can solve for F’ ( x ): '( ) '( ) '( ) g t F x f t = TANGENTS Equation 1
The slope of the tangent to the curve y = F ( x ) at ( x , F ( x )) is F’ ( x ). Thus, Equation 1 enables us to find tangents to parametric curves without having to eliminate the parameter. TANGENTS

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Using Leibniz notation, we can rewrite Equation 1 in an easily remembered form: if 0 dy dy dx dt dx dx dt dt = TANGENTS Equation 2
If we think of a parametric curve as being traced out by a moving particle, then dy / dt and dx / dt are the vertical and horizontal velocities of the particle. Formula 2 says that the slope of the tangent is the ratio of these velocities. TANGENTS

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From Equation 2, we can see that the curve has: A horizontal tangent when dy / dt = 0 (provided dx / dt ≠ 0). A vertical tangent when dx / dt = 0 (provided dy / dt ≠ 0). TANGENTS
TANGENTS This information is useful for sketching parametric curves.

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As we know from Chapter 4, it is also useful to consider d 2 y / dx 2 . This can be found by replacing y by dy/dx in Equation 2: 2 2 d dy d y d dy dt dx dx dx dx dx dt ÷ = = ÷ TANGENTS
A curve C is defined by the parametric equations x = t 2 , y = t 3 – 3 t. a. Show that C has two tangents at the point (3, 0) and find their equations. b. Find the points on C where the tangent is horizontal or vertical. c. Determine where the curve is concave upward or downward. d. Sketch the curve. TANGENTS Example 1

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Notice that y = t 3 – 3 t = t ( t 2 – 3) = 0 when t = 0 or t = ± . Thus, the point (3, 0) on C arises from two values of the parameter: t = and t = – This indicates that C crosses itself at (3, 0). 3 3 TANGENTS Example 1 a 3
Since the slope of the tangent when t = ± is: dy / dx = ± 6/(2 ) = ± 2 / 3 3 3 1 / 2 2 dy dy dt t t dx dx dt t t - = = = - ÷ TANGENTS Example 1 a 3 3 3

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So, the equations of the tangents at (3, 0) are: ( 29 ( 29 3 3 and 3 3 y x y x = - = - - TANGENTS Example 1 a
C has a horizontal tangent when dy / dx = 0, that is, when dy / dt = 0 and dx / dt ≠ 0.

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Chap10_Sec1 - 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES...

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