Calculus AB Homework 4.2: Related Rates

This paragraph introduces a related rates problem involving a sailboat docked near a pulley system. The sailboat is initially at rest, and a rope attached to the bow is pulled in at a rate of 2 feet per second. The goal is to determine how fast the boat is approaching the dock when the rope's length from the bow to the pulley is 13 feet. The problem sets up by drawing a right triangle with a constant height of 5 feet (the height difference between the pulley and the boat's bow) and using the Pythagorean theorem to relate the variables X (the length from the bow to the dock) and Z (the rope's length to the pulley). The rate of change of Z with respect to time (dZ/dt) is given as -2 feet per second, and the task is to find dX/dt when Z equals 13 feet.

The second paragraph presents a related rates problem concerning a baseball player running from second to third base and an umpire standing on home plate. The player advances at a rate of 24 feet per second, and the problem is to find the rate at which the angle (theta) between the third base line and the umpire's line of sight to the runner is changing when the runner is 30 feet from third base. The problem begins by sketching a square baseball diamond and identifying the given rates and unknown rates. The relationship between the variables is established using the tangent function, relating theta to the distance X from the runner to third base. Implicit differentiation is applied to find dTheta/dt when X is 30 feet.

This paragraph discusses a related rates problem about a cone-shaped sand pile formed by sand being dumped from a conveyor belt. The rate of sand dumping is 10 cubic feet per minute, and the challenge is to find how fast the height of the pile is changing when the pile contains 1000 cubic feet of sand. The solution involves using the volume formula for a cone (V = 1/3 * pi * r^2 * h), where the radius is equal to the height (X), and differentiating this equation with respect to time to find dX/dt. The known rate of volume change (dV/dt) is substituted into the equation to solve for the rate of height change when the volume is 1000 cubic feet.

The fourth paragraph describes a scenario where a swimming pool with a length of 60 feet and a width of 25 feet is being filled with water. The pool's depth varies uniformly from 3 feet at the shallow end to 15 feet at the deep end. The rate of water filling is 800 cubic feet per minute, and the problem is to determine the rate at which the depth of water at the deepest point is increasing when the water is 5 feet deep. The solution involves setting up a right triangular prism model for the water in the pool and using the volume formula for a prism to relate the volume (V) to the depth (Y) and the horizontal distance (X). Similar triangles are used to eliminate X and express the volume in terms of Y alone, which is then differentiated with respect to time to find dY/dt when Y equals 5 feet.

In this paragraph, a 25-foot ladder is leaning against a wall, and the base is being pulled away from the wall at a rate of 3 feet per second. The problem is to find how fast the top of the ladder is moving down the wall when the base is 7 feet from the wall. The solution uses the Pythagorean theorem to relate the horizontal distance (X) from the wall and the vertical height (Y) of the ladder. The derivative of the Pythagorean equation with respect to time is taken, and known values are substituted to solve for the rate of change of Y (dy/dt) when X is 7 feet.

The final paragraph presents two related problems involving a ladder leaning against a wall. The first part asks for the rate of change of the area of the triangle formed by the ladder, the wall, and the ground when the ladder's base is 7 feet from the wall. The area formula (A = 1/2 * base * height) is used, and the derivative of the area with respect to time is calculated using the known rates from the previous problem. The second part of the problem asks for the rate at which the angle between the ladder and the wall is changing when the base is 7 feet from the wall. The sine function is used to relate the angle (Theta) to the horizontal distance (X), and the derivative of this relationship with respect to time is calculated to find dTheta/dt.