Advanced Engineering Mathematics
6th Edition
ISBN: 9781284105902
Author: Dennis G. Zill
Publisher: Jones & Bartlett Learning
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Chapter 9.2, Problem 2E
To determine
To graph: The curve and the velocity and acceleration
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11.6: Problem 8
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Jackie and Christine are racing again! We only get some discrete data this time, but we can find the details through calculation! 3 minutes after the race starts, Jackie is 12 miles away from the starting point, and Christine is 27 miles ahead of Jackie. 13 minutes after the race starts, Jackie is 52 miles away from the starting point, but now Christine is only 17 miles ahead from Jackie. Assume they are going at a constant speed. (a) Present their movement on the xy-plane, where the x- and y-axis represent the time since the race starts and their distances from the starting point respectively. Label the four points clearly. (b) Find the speed of Christine and Jackie. You may use the unit “miles per minute” or, if you are comfortable with unit conversion, “miles per hour.” (c) Is anyone given a head start? If so, who is that and how far was that? (d) How far is Jackie from the starting point when he catches up with Christine?
A steel ball weighing 128 pounds is suspended from a spring. This stretches the spring 128/257 feet.
The ball is started in motion from the equilibrium position with a downward velocity of 9 feet per second.The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second) .
Suppose that after t seconds the ball is y feet below its rest position. Find y in terms of t. (Note that this means that the positive direction for y is down.)
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- 2. If a particle moves in the xy-plane so that at any time t > 0, its position vector is (In(1² + 5t), 3r² ), find its velocity vector at time t = 2.arrow_forward3. The position of a particle travelling along a horizontal line in t seconds is s(t) = 2t° – 15t - 84t + 3 meters. (a). Find the velocity and acceleration functions at time t = 2 seconds. (b). Is the particle moving from right to left or from left to right at time t = 7 seconds.arrow_forward- 22 - 2 -2 (3,-4) that has a positive a component. For the function f(x, y) = find a unit tangent vector to the level curve at the pointarrow_forward
- Suppose we do not know the path of a hang glider, but only its acceleration vector a(t) = -(3 cos t)i - (3 sin t)j + 2k. We also know that initially (at time t = 0) the glider departed from the point (4, 0, 0) with velocity v(0) = 3j. Find the glider’s position as a function of t.arrow_forwardEvaluate |A|, where A is the 76 x 76 matrix: 1 2 3 4 76 ... 0 2 3 4 76 ... 0 0 3 4 76 ... A 0 0 0 4 76 ... 0 0 0 0 76 ... Note: The syntax for factorial is just x! ... ...arrow_forwardProblem 2. Suppose f(x, y) = 10e – 7xe", P =(-2,1), and u is the unit vector in the direction of (5, 2). A. Find the gradient of f(x, y) and evaluate at the point P = (-2,1). Vf = (-7(x*e^x+e^x),20*y*e^(y^2)) (Vf) (P) (-7*(2*e^2+e^2),20*e) B. Find the tangent plane to f(x, y) at the point (-2, 1). z = -7*(2*e^2+e^2)*(x+2)+20*e*(y-1)+10*e+14e^(-2) C. Find the differential df and evaluate at the point (-2, 1) df = -7(x*e^x+e^x)*dx+20*y*e^(y^2)*dy %3D df(-2, 1) = -7*(2*e^2+e^2)*dx+20*e*dy Use the differential at the point (-2, 1) to estimate the change in the values of f(x, y) if x changes by -0.5 and y changes by 0.2. df -7*(2*e^2+e^2)*0.5+20*e*0.2 D. Find the directional derivative of f(x,y) at P = (-2, 1) in the direction of u. D,f = -7*(2*e^2+e^2)*5/sqrt(29)+20*e*2/sqrt(29) E. At the point P = (-2,1), find the direction of maximum ascent. Your answer should be a unit vector. u <-7*(2*e^2+e^2)/sqrt(49*(2*e^2+e^2)^2,400*e^2).: Find the maximum rate of change of f(x, y) at P = (-2, 1).…arrow_forward
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