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102 - MATH 153 Selected Solutions for 10.2 Exercise 6 Find...

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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
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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, since x is negative for all t -values in the range, we’ll need to integrate - x .
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