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An object is moving around the unit circle so its x and y coordinates change with time as x=cos(t) and y=sin(t). Assume 0

An object is moving around the unit circle so its x and y coordinates change with time as x=cos(t) and y=sin(t). Assume 0

Algebra and Trigonometry (MindTap Course List)
Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter6: Trigonometric Functions: Unit Circle Approach
Section6.2: Trigonometric Functions Of Real Numbers
Problem 1E: Let Px,y be the terminal points on the unit circle determined by t. Then sin t =____, cos t =_____...
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An object is moving around the unit circle so its x and y coordinates change with time as x=cos(t) and y=sin(t). Assume 0 <t</2. 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 as shown where two corners are at the x and y intercepts and the third one is at the origin. (Note that both the point and the slope of the tangent lin you need for the line equation will depend on t.) The identity sin(2t)=2sin(t)cost(t) might be useful in some parts of this question. (a) The slope of the tangent line to the circle through P(t) is -cot (£) Hint: Your answer should depend on t (b) Now, using the point-slope form, you can write the equation of the tangent line as y-sin(t) -cot(t) Again, your answers should depend on t (c) The area of the right triangle, in terms of t, is a(t)= (d) (x-cos(t) (e) lim a(t)=00 t-pl/2- lim a(t)= ∞ t-ot (f) lim a(t)= (g) With our restriction on t, the smallest t so that a(t)=2 is (h) With our restriction on t, the largest t so that a(t)=2 is (i) The average rate of change of the area of the triangle on the time interval [7/6/4] is (i) The average rate of change of the area of the triangle on the time interval [x/4,x/3] is
Transcribed Image Text:An object is moving around the unit circle so its x and y coordinates change with time as x=cos(t) and y=sin(t). Assume 0 <t</2. 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 as shown where two corners are at the x and y intercepts and the third one is at the origin. (Note that both the point and the slope of the tangent lin you need for the line equation will depend on t.) The identity sin(2t)=2sin(t)cost(t) might be useful in some parts of this question. (a) The slope of the tangent line to the circle through P(t) is -cot (£) Hint: Your answer should depend on t (b) Now, using the point-slope form, you can write the equation of the tangent line as y-sin(t) -cot(t) Again, your answers should depend on t (c) The area of the right triangle, in terms of t, is a(t)= (d) (x-cos(t) (e) lim a(t)=00 t-pl/2- lim a(t)= ∞ t-ot (f) lim a(t)= (g) With our restriction on t, the smallest t so that a(t)=2 is (h) With our restriction on t, the largest t so that a(t)=2 is (i) The average rate of change of the area of the triangle on the time interval [7/6/4] is (i) The average rate of change of the area of the triangle on the time interval [x/4,x/3] is
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