Given the space curve x = t, y = t2, z = 2/3 t3, the curvature is
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Given the space curve x = t, y = t2, z = 2/3 t3, the curvature is
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- A spaceship travels through deep space to a space station and needs to make a heading correction to get around a nebula. The captain orders the ship to travel 3.1 x 106 kilometers before turning 70 ° and to travel 2.5 x 106 kilometers in that direction. If the captain had not ordered a heading correction, what would have been the magnitude (in kilometers) and the direction (in degrees counterclockwise from the + x-axis) (Counter clockwise: counterclockwise) of the trajectory of the spaceship if it had traveled through the nebula?Solve the quadratic equation 3.00t2 − 4.00t − 6.00 = 0 using the quadratic formula, finding both solutions.If you project a tennis ball at speed v0 on a horizontal surface, what must the angle of the projection be to get the maximum area under the trajectory? What is the area? Hint: A = ∫ydx
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