Problem 1. Explain how you know when to use the power rule, product rule, quotient rule, and chain rule to calculate derivatives. What do functions look like that require each of these rules? Be specic! Problem 2. Let f(:12) = sinx and g(:c) = J}. (a) Find 11(13) = f (g(a:)) and calculate h'($). (b) Find k(:v) = g( f (32)) and calculate k'(:c). Problem 3. Find the derivatives of the following functions. (a) g(x) = (5x - 8) - (2 + 4 ) = - 2+ 4 5 x - 8 Two methods: (1)product +chain rule or (2) quotient rule-pros/cons? (b) h(ze) = sin (x2 + 5x) (c) f(x) = sec( V4x + 1) (d) g(a) = V2ac2 + 5x + 4 (e) p(x) = sin (a2 cos ac )Problem 4. Particle motion is described by several phrases that are very similar, but have important differences. Work with your classmates to compare and contrast the following phrases and to describe the di'erence: (a) \"the particle has a speed of 5 m/s\" vs. \"the particle has a velocity of 5 m/ s\" (b) \"the particle is speeding up\" vs. \"the particle has positive acceleration\" (c) \"displacement\" vs. \"total distance traveled\" Problem 5. The graph to the right shows the velocity v(t) of a particle moving on a horizontal coordinate line (in m/s). Use this graph to answer the following questions. Your answers should be in interval notation. (a) When is the particle moving in the positiveegative direction? (b) What is the particle's maximum speed? velocity? (c) When is the particle motionless? (d) When is the acceleration positive? Negative? (e) When is the particle speeding up/slowing down? (f) At t = 9 the particle: {circle all that apply) I. is moving in the positive direction 11. is moving in the negative direction III. is speeding up IV. is slowing down V. has positive acceleration VI. has negative acceleration VII. is at position y=-1.5 Problem 6. A ball is thrown vertically upward with an initial velocity of 10 meters per second. Its height after 15 seconds is given by h(t) = at2 + 10t + 1 (a) Find the value of a if the ball reaches its maximum height after 5 seconds. (b) What is the maximum height the ball reaches? (c) When will the ball hit the ground? (d) How fast is the ball traveling when it hits the ground? Problem 7. The Mackinac bridge spans the strait of Mackinac and provides a connection between the upper and lower peninsula of Michigan. Its total length is 8038.2 meters, its main span is 1158 meters and its towers are about 140 meters tall when measured from the bridge deck. You might find this Desmos graph useful: https://www.desmos.com/calculator/bgho3sq0ar Note the picture is not an exact match because the bridge is not being photographed perfectly in profile (but it is pretty close!) Suspended Length Main Span Length - & PIER - & TOWER & BRIDGE - & TOWER -& PIER SOUTH ANCHORAGE ANCHORAGE 155. 135 -1 471' -3" 3800 ' -0" (471' -3" 18 SOUTH SIDE SPAN 19 20 NORTH SIDE SPAN 21) 135' -0" PIER PIER PIER PIER PIER PIER 1. If we assume the main suspension cable forms a parabola, find an equation f(x) that represents its position above the bridge deck. Note that the cable touches the bridge deck in the middle of the main span and goes to the top of each tower of the main span. Hint: You can choose coordinates so that the midpoint of the main span is the origin 2. As we can see below, the second derivative of the cable f"(x) is related to the density of the roadway pr (the mass per unit length), the acceleration due to gravity g and the tension in the cable T' by the following formula f"(2) = 9Pr T If the roadway has a density of 115, 620 kg/m, how much tension must the cable be able to withstand? Hint: your answer should have units (kg m/s2).The rest of this question is devoted to understanding why f\"(:c) should be equal to 91% and is very challenging. It uses material from outside of class, namely: free body diagrams, force balancing and the small angle approximation. Consider the segment of cable w and a: + h. It has three forces acting on it: the tension from the left T1, the tension from the right T2 and the weight of the roadway FT it is supporting. ( h l x x+h Because the bridge is not in motion these forces must be balanced in each direction. Assume 61 is the angle the cable makes with the horizontal at a: and 92 is the angle the cable makes with the horizontal at :c + h. Then the forces being balanced means T1 cos(91) = T2 cos(02) and T1 sin(91) T2 sin(92) = Fr. . The small angle approximation says cos(9) 2 1 for 6 small enough. Assuming this applies to our work, what do we know about T1 and T2? How does this relate to the formula in part 2? . Recalling that pr is the mass per unit length of the roadway, what is a formula for the weight of the roadway supported by this section of cable? Hint: your answer should have units of (kg m/s2) . Using the small angle approximation sin(0) z 0 z tan(6) for 6 small, nd an expression relating f'(w) and sin(91). Do the same for f'($ + h) and sin(32).Hz'nt: Find a way to relate tan(01) to the slope of the cable at :c, and note that f' (as) is also the slope of the cable. . Plug the previous two parts into T1 sin(61) T2 sin(02) = Fr to nd an expression relating f ' (:c + h), f' (as), T, h, g and pr. . Rearrange the expression you found above so that you can take the limit as h > 0. Then do so, to end up with an expression involving f\"(:r). Hint: your answer should be the same as the formula from part 2