Calculus travel time question?
Let x be the distance along the riverbank the hunter travels before heading in a straight line for his cabin. The distance he will then travel over the rocky terrain is then the hypotenuse, h, of a right triangle with sides of length 3 and (8 - x) miles, (this should be clear once you've drawn a diagram representing this scenario). So h = sqrt(3^2 + (8 - x)^2) = sqrt(9 + 64 - 16x + x^2) = sqrt(x^2 - 16x + 73). Now what we need to minimize is T, the time of travel. Since time = distance/speed, we have T = (x/5) + (h/2) = (x/5) + (1/2)*sqrt(x^2 - 16x + 73). Now find dT/dx and set it equal to 0 to find the critical points: dT/dx = (1/5) + (1/2)*(1/2)*(x^2 - 16x + 73)^(-1/2) * (2x - 16) = (1/5) + (1/2)*(x - 8)*(x^2 - 16x + 73)^(-1/2) = 0 when 1/5 = (1/2)*(8 - x)*(x^2 - 16x + 73)^(-1/2) ----> (x^2 - 16x + 73)^(1/2) = (5/2)*(8 - x). Now square both sides to get x^2 - 16x + 73 = (25/4)*(8 - x)^2 ---> 4*(x^2 - 16x + 73) = 25*(64 - 16x + x^2) ----> 4x^2 - 64x + 292 = 25x^2 - 400x + 1600 ---> 21x^2 - 336x + 1308 = 0 ----> 7x^2 - 112x + 436 = 0. Now use the quadratic formula to solve for x: x = (112 +/- sqrt((-112)^2 - 4*7*436))/(2*7) = (112 +/- sqrt(336))/14 = (112 +/- 4*sqrt(21))/14 = 8 +/- (2/7)*sqrt(21). Now clearly x <= 8, so x = 8 - (2/7)*sqrt(21) = 6.69 miles to 2 decimal places. Edit: To make this solution complete, I should note that d/dx(dT/dx) = 9/(x^2 - 16x + 73)^(3/2) > 0 for x = 6.69 miles, (for all x, in fact), so by the second derivative test we can conclude that T is minimized when x = 8 - (2/7)*sqrt(21) = 6.69 miles. (I've spared you the details of calculating the second derivative, but you can do it yourself to confirm.)
Calculus BC Question?
since acceleration is directly proportional to time, we have the following: a(t) = -kt v(t) = integral of a(t) v(t) = integral of (-kt) v(t) = -kt^2/2 + C (at t = 0, v = 300) 300 = C v(t) = -kt^2/2 + 300 at t = 10, v(t) = 0 0 = -50k + 300 50k = 300 k = 6 v(t) = -3t^2 + 300 (part a) integral (from 0 to 10) (absolute value of v(t)) = total distance traveled since v(t) is positive on interval from 0 to 10, we can do: integral(0 to 10)(v(t)) = total distance traveled total distance = integral(0 to 10)(-3t^2 + 300) total distance = [-t^3 + 300t](0 to 10) total distance = [-1000 + 3000] - [0] total distance = 2000 voila
Velocity/Time question: Calculus?
f(t) = -16t² + 64t = -16t(t - 4) = 0 t = 0 or 4. Reject t = 0 because the problem says the pomegranate was thrown up. So it must be t = 4sec to reach the ground. f(t) = -16(t² - 4t) f'(t) = -32t + 64 = -32(t - 2) = 0 t = 2 when it reaches the peak of thrown f(2) = -16(2² - 4(2)) = -16(4 - 8) = 64feet Yin
AP Calculus AB hard particle motion question?
particle X moves along the positive x-axis so that its position at time t≥0 is given by x(t)=5t³-9t²+7. (A) Is particle X moving toward the right or toward the left at time t=1? Give a reason for your answer. (B) At what time t≥0 is particle X farthest from the left? Justify your answer. (C) A second particle, Y, moves along the positive y-axis so that its position at time t is given by y(t) = 7t +3. At any time t, t≥0, the origin and the positions of the particles X and Y are the vertices of a triangle in the first quadrant. Find the rate of change of the area of the triangle at time t=1. Show the work that leads to your answer.
Doubling time question! pre calculus help!!?
Profits in the nth year will be P(o) * 1.13^n and we want profits to be 2P(o) for doubling 2P(o) = P(o) * 1.13^n 2 = 1.13^n take log of both sides: ln 2 = ln (1.13^n) = n ln 1.13 n = ln 2 / ln 1.13 = 5.67 years doubling time: 5.67 years check: 1.13^5.67 = 1.99965 (we'll call it 2, good to 2 decimal places)
Calculus, velocity/time/distance question?
You got (a) right. So moving on and accepting your statement that you need to "use calculus." Set up the distance equation from t=0 (when the brakes were first applied and you know your distance to the bear) to t=12.5 s (when the car has stopped.) In calculus form, it looks like this: 1. D = ʃ V(t) dt, from t=0 to 12.5s But you must now figure out V(t). It is: 2. V(t) = 90 km/hr - 2 m/s • t = (25 - 2•t), putting 90 km/hr as 25 m/s So plug that into (1) above: 3. D = ʃ (25 - 2•t) dt, from t=0 to 12.5s 4. D = ( 25 ʃ dt - 2 ʃ t dt ), from t=0 to 12.5s 5. D = ( 25•t - t² ), from t=0 to 12.5s 6. D = [ ( 25•12.5 - 12.5² ) - ( 25•0 - 0² ) ] 7. D = ( 25 - 12.5 ) • 12.5 8. D = 12.5² = 156.25 meters So that is 106.25 meters past the bear. Just to double-check your work, the average velocity given a smooth deceleration was 12.5 m/s (halfway between 25 m/s and 0 m/s.) You already know how long. So 12.5 m/s times 12.5 s does in fact match up with the results in (8) above. Which is a good thing.
Calculus question: velocity and distance?
The acceleration function a(t) (in m/s2) and the initial velocity v(0) are given for a particle moving along a line. a(t)=2t+2, 0