221, Math Fall 2006, Lecture 7, Solutions to Midterm 2. Try to do first the problems that look simplest to you. You don't have to follow the order. Avoid spending too much time on one question with others not done yet. Good luck! 1. Define what is the inverse function and what is the condition for its existence. State the formula for the derivative of inverse function. Evaluate cos(sin-1 (1/2) - cos-1 (1/2)). The inverse function f -1 exists if and only if f is one to one. If b is in the range of f and f (a) = b, then f -1 (b) = a. Recall that cos( - ) = cos cos - sin sin . Also, cos(sin-1 1/2) = 1 - (sin(sin-1 1/2))2 = 1 - (1/2)2 = 3/2. Similarly, sin(cos-1 1/2) = 3/2. Then cos(sin-1 (1/2) - cos-1 (1/2)) = cos(sin-1 (1/2)) cos(cos-1 (1/2))+ sin(sin-1 (1/2)) sin(cos-1 (1/2)) = ( 3/2)(1/2) + (1/2)( 3/2) = ( 3/2).
2. A particle moves along the hyperbola y = 1/x in the first quadrant in such a way that its x-coordinate (measured in meters) increases at a steady 3m/sec. How fast is the angle of inclination of the line joining the particle to the origin changing when x = 1m? Observe that tan (t) = y(t)/x(t). Notice also that since the particle moves over hyperbola, y(t) = 1/x(t). Differentiating, we get 1 d 1 2x (t) (t) = =- . 2 2 (cos((t)) dt x(t) x(t)3 When x = 1, we also have y = 1/x = 1, and so tan = 1, = /4. Plugging in all the numbers, we get 2 (t) = -6, so (t) = -3 radians per second. 3. Assume 0 < t < /2. Find an equation for the line tangent to the curve x(t) = tan t, y(t) = 1 + cos t at a point corresponding to t = /6. Can you describe the curve above in terms of the Cartesian equation? Compute x (t) = 1/(cos t)2 , y (t) = 1 - sin t. Then dy dx dy = / = (1 - sin t)(cos t)2 . dx dt dt dy Putting in t = /6, we arrive at dx = (1 - 1/2) 3 = 3/8. 4 Also when t = /6, x = 1/ 3 and y = 1 + 3/2. Therefore the equation of tangent is L(x) = 1 + 3/2 + 3/8(x - 1/ 3).
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Finally, if 0 t , then tan t = 1 - (cos t)2 cos t. Thus x = gives a Cartesian expression for the curve.
1 - (y - 1)2 /(y -1)
dy dx
4. Assume an implicit function y(x) is defined by ln y + xy = x Find point (1, 1). Let us differentiate, using chain rule: y -1 y + y + xy = 1, so y = In particular, y (1) = 0. Next, y = (y - 2yy )(1 + xy) - (y - y 2 )(xy + y) . (1 + xy)2 y - y2 . 1 + xy
and
d2 dx2
at y a
Putting in y = 0, x = y = 1 leads to y = 0. 5. State and prove Rolle's theorem. You may assume that if the function f is differentiable and x0 is its local maximum or minimum, then f (x0 ) = 0. Rolle's Theorem: Assume that f (x) is defined on [a, b], differentiable on (a, b) and continuous on [a, b]. Assume that f (a) = f (b). Then there exists c in (a, b) such that f (c) = 0. Proof: If f (x) is constant, then the derivative is zero at every point, and theorem holds. So we assume that f is not constant. In that case there should exist a point c in (a, b) where f takes its maximum or minimum value different from f (a) and f (b). We know that a differentiable function has derivative equal to zero at the point where it takes maximum (or minimum) value. This completes the argument.
1 6. Investigate how many zeroes the function f (x) = sec x - x3 + 5 has on the interval [0, /2]. Explain your reasoning.
Observe f becomes very large negative when x approaches 0 because of -1/x3 term. Also, f becomes very large positive as x approaches /2 because of sec x term. By the Intermediate Value Theorem, there must be at least one zero. On the other hand, 3 f (x) = sec x tan x + 4 > 0 x for any x in our interval. If there were more than one zero of f in our interval, by Rolle's Theorem we could find a point x0 between them where f (x0 ) = 0. But the derivative is positive, so such point cannot exist, and therefore there cannot be more than one zero. 7. Find absolute maximum and minimum of the following functions: a. f (x) = 2x - 2 sin 2x on [-/2, /2];
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b. f (x) = 2x3 - 8x2 + 8x - 7 on [0, 3]. a. Find the derivative: f (x) = 2 - 2 2 cos 2x. The function is odd, so it suffices to look at x 0. The derivative vanishes if cos 2x = 1/ 2, 2x = /4. Thus or x = /8 are the critical points. Now f (/8) = /4 - 2/ 2 = /4 - 1. Also, f (-/8) = -/4 + 1. We have to check the endpoints: f (/2) = - 2 sin = and f (pi/2) = -. Clearly, f (/2) is the absolute maximum and f (-/2) is the absolute minimum. b. f (x) = 6x2 - 16x + 8 = 2(3x2 - 8x + 4). Finding the roots of the quadratic 8 gives critical points x = 2, 2/3. Observe that f (2) = -7 and f (2/3) = -4 27 . Now at the endpoints f (0) = -7 and f (3) = -1. Clearly, x = 3 is the absolute maximum, while x = 0 and 2 are two absolute minima.