Mean Value Theorem. is continuous at every point of the closed interval,

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1 Mean Value Theorem The Mean Value Theorem connects the average rate of change (slope of the secant between two points [a and b]) with the instantaneous rate of change (slope of tangent at some point c). The theorem says that somewhere between a and b on a differentiable curve, there is at least one tangent line parallel to the slope of the secant between points a and b. Definition: Mean Value Theorem y f is continuous at every point of the closed interval,, f b f a ab, at which fc If ab and differentiable at every point of its interior ab, then there is at least one point c in b a Physically, the Mean Value Theorem says that the instantaneous change at some interior point must equal the average change over the entire interval. a) Show that 0,1 f 1 satisfies the Mean value Theorem on the interval b) Find the value c in (0, 1) that satisfies the equation. a) Since f is continuous on 0,1 and differentiable (0, 1). Since f 0 1 and f 1, the Mean Value Theorem guarantees a point c in the interval (0, 1)

2 b) fc f 1 f c 1 c 1 1 c The function f, 0 1 0, 1 Is zero at =0 and at =1. Its derivative is equal to 1 at every point between 0 and 1, so f is never zero between 0 and 1, and the graph of f has no tangent parallel to the chord from (0, 0) to (1, 0). Eplain why this does not violate the Mean Value Theorem. Because the function is not continuous on 0,1 ; the function does not satisfy the hypothesis of the Mean Value Theorem and need not satisfy the conclusions. The geometric mean of any two positive number a and b is ab. Show that if f on any interval [a, b], then the value of c in the conclusion of the Mean Value theorem is c c ab f b f a b a b a b a ab f c c ab c ab c ab Positive since a and b are both positive, and c is between a and b so it also must be positive.

3 Suppose f is differentiable in (a, b) and f M for all a, b a) Prove that f f M for all, a, b b) Prove sin a a, a 1 1 c) Using a), prove sin sin a) From Mean Value theorem f f 1 1 f f 1 1 f M M M f f M 1 1 b) a a a f c, c a,0 sin sin sin 0 sin sin 0 a 0 sin sin 0 sin sin sin a a a a a a a 0 1 cos a c f c c) sin cos 1 M sin M M sin f f M 1 1 M 1 1 M 1 1

4 Definition: Increasing and Decreasing Functions Let f be continuous on If 0 If 0 ab, and differentiable on ab, f at each point of (a, b), then f increases on [a, b] f at each point of (a, b), then f decreases on [a, b] Corollary: Functions with the same derivative differ by a constant Prove that has eactly one solution. 3 4 is always positive, then the function is always increasing. Since the derivative Therefore the function cannot have two solutions as it does not have a change of direction (does not decrease). Definition: Antiderivative A function F() is an antiderivative of a function f() if F f for all in the domain of f. The process of finding the Antiderivative is anti-differentiation.

5 4 4 Find the Antiderivative of f F C C 5 3 Find the Antiderivative of f cos 3 F sin 3 C f 10e e 5 3 Find the Antiderivative of F e e C e e C 3 f 3sin 3 e Find the Antiderivative of F cos 3 e C 3 3

6 On the moon, the acceleration due to gravity is 1.6 sec m a) If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 0 seconds later. b) How far below the point of release is the bottom of the crevasse? c) If instead of being released from rest, the rock is thrown into the crevasse from the m, when will it hit the bottom and how same point with a downward velocity of 3 sec fast will it be going when it does? a) Since acceleration the derivative of the velocity, then vt 1.6 Now finding the Antiderivative will find the velocity function v(t). 1.6 v t t C We need to find the value of c. We know that when t=0, v(t)=0 v C 00C C 0 Therefore vt 1.6t and v Thus the rock will be travelling at 3 m/sec. b) Let st represent the position of the rock at any time t. Since st vt 1.6t s t t C 0.8 Now s0 0, therefore C=0, so st 0.8t s m Thus

7 c) As in above v t 1.6t C v C 3 C Therefore v t 1.6t 3 s t 1.6t 3 s t 0.8t 3t C s(0) C 0 C s t 0.8t 3t Now we know that the depth is 30 metres Therefore t 3t 0.8t 3t 30 0 t 1.963, t The velocity is v This tells us that the rock will hit the bottom after about seconds and will hit with velocity of about m/sec A ball is dropped from a height 45 m above the ground. At the same instant another ball is thrown downward with a velocity of 0 m/s from a height of 60m above the ground. Which ball hits the ground first given that the acceleration due to gravity is 10 m/s. In this question, we choose down to be negative. Therefore my acceleration will be negative, and the tossed ball with have a negative velocity. Let s find the velocity and position functions of the dropped ball. a t 10 v t 10t C C C 0 v t 10t

8 s t 5t C C C 0 s t 5t Now let s find the time when the position is minus t 9 t t 3 [omit -3] Let s find the velocity and position functions of the thrown ball. a t 10 v t 10t C C C 0 v t 10t 0 s t 5t 0t C C C 0 s t 5t 0t Now let s find the time when the position is minus t 0t 5t 0t 60 0 t 4t t [omit -6] t t Time difference is 3-=1 Therefore the thrown ball will hit the 1 second before the dropped ball.