# 101 - MATH 153 Selected Solutions for §10.1 Exercise 12...

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Unformatted text preview: MATH 153 Selected Solutions for §10.1 Exercise 12: Eliminate the parameter to ﬁnd a Cartesian equation of the curve, then sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases. x = 4 cos θ, y = 5 sin θ, −π/2 ≤ θ ≤ π/2 Solution: As this parametrization is similar to that of a circle, we use the same trigonometric identity as before: y x sin2 θ + cos2 θ = 1 ⇔ + =1 5 4 x2 y 2 + = 1. = 25 16 You may recall that this is the equation of an ellipse, with axes of lengths 4 and 5. As θ only traces out half a period of sine and cosine, we’ll only get half of the ellipse: 4 2 1 2 3 4 2 4 Figure 1. A sketch of x = 4 cos θ, y = 5 sin θ. 1 2 Exercise 42: If a and b are ﬁxed numbers, ﬁnd parametric equations for the curve that consists of all possible positions of the point P in the ﬁgure (see book), using the angle θ as the parameter. The line segment AB is tangent to the larger circle. Solution: It should be fairly obvious, after dropping an altitude for the radius at angle θ of the inner circle, that the y -coordinate of the point is just y = b sin θ. Looking at the triangle OAB , which is a right triangle with right angle A, we see that cos θ = a/x, or that x = a sec θ. Thus, a combined parametrization for the possible points P would be x = a sec θ, y = b sin θ, 0 ≤ θ ≤ 2π. 2 1 4 2 2 4 1 2 Figure 2. A sketch for a = 2, b = 1 (the black curve). ...
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101 - MATH 153 Selected Solutions for §10.1 Exercise 12...

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