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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 /2. At a given time t, the tangent line to the unit circle at the position A(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 line 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. 1 (a) The slope of the tangent line to the circle through A(t) is 2 sin (x) cos(x) X Hint: Your answer should depend on t. (b) Now, using the point-slope form, you can write the equation of the tangent line as Again, your answers should depend on t. (c) The area of the right triangle, in terms of t, is a(t)= (d) lim a t - pl/2 (e) lim a(t)= (F) lim a(t)= (g) With our restriction on t, the smallest & 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 [1/6,7/4] is j) The average rate of change of the area of the triangle on the time interval [>4,1/3] isThe solutions (x,y) of the equation x2 + 16yz = 16 form an ellipse as pictured below. Consider the point P as pictured, with x-coordinate 1. (a) The ellipse is not given by a single function as it fails the vertical line test. It is given by two function. The points P and Q shown are in the upper part which can be described as y=f(x) where f(x)= 1 - 16 (b) Let h be a small non-zero number and form the point Q with x-coordinate 1+h, as pictured. Then, the y-coordinate of Q is given by 1 - (1 + h)2 16 (c) Let h be a small non-zero number and form the point Q with x-coordinate 1+h, as pictured. The slope of the secant line through PQ, denoted s(h), is given by the formula (d) Rationalize the numerator of your formula in (c) and simplify 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 have finished your simplification, neither the numerator nor the denominator should become zero when h=0.) When you do this f(h)= -4-h X g(h)= (e) The slope of the tangent line to the ellipse at the point P is lim s (h) h-+0 4V3 XThe solutions (x,y) of the equation x + 16y = 16 form an ellipse as pictured below. Consider the point P as pictured, with x-coordinate 1. (a) The ellipse is not given by a single function as it fails the vertical line test. It is given by two function. The points P and Q shown are in the upper part which can be described as y=f(x) where f(x)= 1 16 (b) Let h be a small non-zero number and form the point Q with x-coordinate 1+h, as pictured. Then, the y-coordinate of Q is given by 1 - (1 +h)2 16 (c) Let h be a small non-zero number and form the point Q with x-coordinate 1+h, as pictured. The slope of the secant line through PQ, denoted s(h), is given by the formula (d) Rationalize the numerator of your formula in (c) and simplify 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 have finished your simplification, neither the numerator nor the denominator should become zero when h=0.) When you do this f ( h ) = X g(h)= (e) The slope of the tangent line to the ellipse at the point P is lim s (h) h-+0 X