102 - MATH 153 Selected Solutions for § 10.2 Exercise 6 :...

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Unformatted text preview: MATH 153 Selected Solutions for § 10.2 Exercise 6 : Find an equation of the tangent to the curve at the point corre- sponding to the given value of the parameter. x = cos θ + sin 2 θ, y = sin θ + cos 2 θ, θ = 0 . Solution : When θ = 0 we are at the point (1 , 1). dy dx = dy/dt dx/dt , so we calculate the two derivatives: dy/dt = cos θ- 2 sin 2 θ, dx/dt =- sin θ + 2 cos 2 θ Plugging in θ = 0, we get dy/dt = 1 , dx/dt = 2 ⇒ dydx = 1 / 2 . Having the point and the slope, we plug into point-slope form: y- 1 = 1 / 2( x- 1) .- 1.5- 1.0- 0.5 0.5 1.0 1.5 2.0- 2.0- 1.5- 1.0- 0.5 0.5 1.0 1.5 Figure 1. The parametric curve and its tangent line at θ = 0. 1 2 Exercise 32 : Find the area enclosed by the curve x = t 2- 2 t , y = √ t and the y-axis.- 1.0- 0.8- 0.6- 0.4- 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 Figure 2. The area in question Solution : We integrate from one root of x to the other; that is, from t = 0 to t = 2. When t = 0 y = 0 and when t = 2 y = √ 2. Also note that, since2....
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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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102 - MATH 153 Selected Solutions for § 10.2 Exercise 6 :...

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