103 - MATH 153 Selected Solutions for §10.3 Exercise 18...

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Unformatted text preview: MATH 153 Selected Solutions for §10.3 Exercise 18: Identify the curve by finding a Cartesian equation for the curve. r = 2 sin θ + 2 cos θ Solution: 2.0 1.5 1.0 0.5 0.5 1.0 1.5 2.0 Figure 1. A graph of r = 2 sin θ + 2 cos θ, with 0 ≤ θ ≤ π . Recall that x = r cos θ and y = r sin θ, so that y x cos θ = . sin θ = , r r 2 Plugging these in gives r = 2x/r + 2y/r, or r = 2x + 2y . But r2 = x2 + y 2 , so we get the equation x2 + y 2 = 2x + 2y , or in standard form (x − 1)2 + (y − 1)2 = 2, √ the equation for a circle of radius 2 and center (1, 1). 1 2 Exercise 34: Sketch the curve with the given polar equation. r = 1 − 3 cos θ 2 1 4 3 2 1 1 2 Figure 2. The curve r = 1 − 3 cos θ. Exercise 44: Sketch the curve with the given polar equation. r2 = cos 4θ 1.0 0.5 1.0 0.5 0.5 1.0 0.5 1.0 Figure 3. The curve r2 = cos 4θ. 3 Exercise 60: Find the slope of the tangent line to the given polar curve at the point specified by the value of θ. r = cos(θ/3), θ=π Solution: 0.5 1.0 0.5 0.5 1.0 0.5 Figure 4. The curve r = cos(θ/3) with its tangent line at θ = π . We use the standard formula to find the slope: dy dy/dθ = dx dx/dθ dr sin θ + r cos θ dθ = dr cos θ − r sin θ dθ (− sin(θ/3)/3) sin θ + cos(θ/3) cos θ = . (− sin(θ/3)/3) cos θ − cos(θ/3) sin θ Plugging in θ = π gives √ − 63 (0) + 1 (1) dy 2 =√ = 3 1 dx − 6 (1) − 2 (0) 1 2 √ −3 6 √ = − 3. Since the point (r, θ) = (1/2, π ) corresponds to the point (x, y ) = (−1/2, 0), we plug into point-slope form to get the equation √ y = − 3(x + 1/2). ...
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This note was uploaded on 07/26/2011 for the course MATH 153 taught by Professor Rempe during the Spring '08 term at Ohio State.

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103 - MATH 153 Selected Solutions for §10.3 Exercise 18...

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