1. How fast does the radius of a spherical soap bubble change when you blow air into it at the rate of 15 cubic centimeters per second? Our known rate is dV/dt, the change in volume with respect to time, which is 15 cubic centimeters per second. The rate we want to find is dr/dt, the change in the radius with respect to time. Remember that the volume of a sphere is V=4/3 pi r^3.
2. A baseball diamond is a square 90 feet on a side. A player runs from first base to second base at a rate of 15 feet per second. At what rate is the player's distance from third base decreasing when the player is halfway between first and second base? We've already set part of this problem up. If we let x be the distance between the player and second base,and y be the distance between the player and third base, then dx/dt=-15 feet per second, and dy/dt will tell us what we want to know. Use the picture to find a relationship that will help you answer the question.
3. A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate is the tip of his shadow moving? We've already set this up part of the way. We know that dx/dt=-2 meters per second, and we're looking for dv/dt . Use the picture to help you find the relationship between x and y, and use it to answer the question asked here.
4. Here's one we haven't worked with before: A circular oil slick of uniform thickness is caused by a spill of 1 cubic meter of oil. The thickness of the oil is decreasing at the rate of 0.1 cm/hr as the slick spreads. (Note: 1 cm = 0.01 m.) At what rate is the radius of the slick increasing when the radius is 8 meters? (You can think of this oil slick as a very flat cylinder; its volume is given by V = AP Calculus AB Semester 1 r2h, where r is the radius and h is the height of this cylinder.)
Never mind! I solved them all and got them right. :)
1. To find the rate of change of the radius of a spherical soap bubble, we need to differentiate the volume formula with respect to time. The volume formula of a sphere is V = (4/3)πr^3, where V is the volume and r is the radius. We are given that dV/dt = 15 cubic centimeters per second.
To find dr/dt, we differentiate the volume formula with respect to time:
dV/dt = (d/dt)(4/3)πr^3
15 = 4πr^2(dr/dt)
dr/dt = 15 / (4πr^2)
2. To find the rate at which the player's distance from third base is decreasing, we need to relate the variables x, y, and their rates of change. Let x be the distance between the player and second base and y be the distance between the player and third base. We are given that dx/dt = -15 feet per second.
By using the Pythagorean theorem, we can establish a relationship between x, y, and their rates of change:
x^2 + y^2 = (90/2)^2
Differentiating both sides with respect to time,
2x(dx/dt) + 2y(dy/dt) = 0
Solving for dy/dt, we get:
dy/dt = -(x/y)(dx/dt)
3. To find the rate at which the tip of the man's shadow is moving, we need to relate the variables x, y, and their rates of change. Let x be the distance the man is from the streetlight (horizontal distance), and y be the length of the shadow. We are given that dx/dt = -2 meters per second (as the man is walking towards the streetlight).
By considering similar triangles formed by the man, the shadow, and the streetlight, we can establish a relationship between x, y, and their rates of change:
(x + y)/y = 5/2
Differentiating both sides with respect to time,
[(dx/dt) + (dy/dt)]/y = 0
Solving for dy/dt, we get:
dy/dt = -(dx/dt)(y/(x + y))
4. To find the rate at which the radius of the oil slick is increasing, we need to relate the variables r, h, and their rates of change. Let r be the radius of the oil slick and h be the height (thickness) of the oil slick. The volume formula of the cylinder is V = πr^2h, where V is the volume.
Given that the thickness of the oil slick is decreasing, dh/dt = -0.1 cm/hr. To find the rate at which the radius is increasing, we need to determine dr/dt when r = 8 meters.
Using the volume formula, we can express h in terms of r:
V = πr^2h
1 = π(8^2)h
h = 1 / (64π)
Differentiating both sides with respect to time,
(dV/dt) = (d/dt)(πr^2h)
0 = 2πrh(dr/dt) + πr^2(dh/dt)
Solving for dr/dt when r = 8 meters,
0 = 2π(8)(1 / (64π))(dr/dt) + π(8^2)(-0.1)
Simplifying,
0 = (16/64)(dr/dt) - 0.8π
dr/dt = 0.8π * (64/16)
dr/dt = 3.2π meters per hour