At a given time t the tangent line to the unit circle

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. At a given time t , the tangent line to the unit circle at the position P(t) will determine a right triangle in the first quadrant. (Connect the origin with the y -intercept and x -intercept of the tangent line.) The identity might be useful in some parts of this question. (a) The area of the right triangle is a(t) = 12 sin ( t ) cos ( t ) . (b) If the tangent line to y = f ( x ) at ( 5 , 4 ) passes through the point (0, 3 ), find f ( 5 ) and f' ( 5 ). f ( 5 ) = 4 f' ( 5 ) = 15 sin(2t)=2sin(t)cost(t)
(c) (d) 1 (e) With our restriction on t , the smallest t so that a(t)=2 (f) With our restriction on t , the largest t so that a(t)=2 (g) The average rate of change of the area of the triangle on the time interval [ π /6, π /4] is . (g) The average rate of change of the area of the triangle on the time interval [ π /4, π /3] is a ( t )= lim t pi /2 a ( t )= lim t 0 + a ( t )= lim t pi /4 is is .
6. 12/12 points | Previous Answers The solutions (x,y) of the equation x 2 + 16 y 2 = 16 form an ellipse as pictured below. Consider the point P as pictured, with x -coordinate 2 . (a) Let h be a small non-zero number and form the point Q with x -coordinate 2 +h , as pictured. The slope of the secant line through PQ , denoted s(h) , is given by the formula $$( ( h 2 4 h +12) −√ 12)4 h . (b) Rationalize the numerator of your formula in (a) to rewrite the expression so that it looks like f(h)/g(h) , subject to these two conditions: (1) the numerator f(h) defines a line of slope -1, (2) the function f(h)/g(h) is defined for h=0 . When you do this f(h)= h 4 g(h)= 4 12 4 h h 2+8 3 . (c) The slope of the tangent line to the ellipse at the point P is = −√ 312

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