3. The radius r of a sphere is increasing at a constant rate of 0.04 centimeters per second. (Note: The volume of a sphere with radius r is v=4/3pir^3 ).
a. At the time when the radius of the sphere is 10 cm, what is the rate of increase of its volume?
b. At the time when the volume of the sphere is 36pi cubic centimeters, what is the rate of increase of the area of a cross section through the center of the sphere?
c. At the time when the volume and the radius of the sphere are increasing at the same numerical rate, what is the radius?
dV/dt = 4πr^2 dr/dt
a) when r=10 and dr/dt = .04
dV/dt = 4π(100)(.04) = ...
b) when V = 36π
36π = (4/3)πr^3
27 = r^3
r = 3
cross section is a circle, A = πr^2
dA/dt = 2πr dr/dt
= 2π(3)(.04)
= ..
c) dV/dt = 4πr^2 dr/dt, but dr/dt = dV/dt = .04
.04 = 4πr^2 (.04)
4πr^2 = 1
r^2 = 1/(4π)
r = 1/(2√π) = ...
So confusing
The radius of an air bubble is increasing at what rate is the volume of the bubble increasing when the radius is 1 centimetre?
The volume of sphere is increasing at the rate of 1200c.cm/sec. the rate increase in its surface area when the radius is 10cm is
a. Well, to find the rate of increase of the volume, we need to differentiate the volume equation with respect to time. So, let's get our math hats on and do some calculus!
The volume of a sphere is given by V = (4/3)πr^3, where r is the radius. Differentiating both sides with respect to time (t), we get:
dV/dt = (4/3)π * 3r^2 * dr/dt
Since dr/dt is given as 0.04 cm/s, and we want to find the rate of increase of volume when the radius is 10 cm, we can plug in these values:
dV/dt = (4/3)π * 3(10)^2 * 0.04 cm/s
Simplifying, we find that the rate of increase of volume is approximately:
dV/dt = 320π cm^3/s
b. Now, we need to find the rate of increase of the area of a cross section through the center of the sphere when the volume is 36π cubic centimeters. Let's call the area A.
Since we know the volume V, we can express it in terms of the radius r using our volume equation:
V = (4/3)πr^3
Plugging in the given volume of 36π cubic centimeters, we have:
36π = (4/3)πr^3
Simplifying further, we find:
r^3 = 27
Taking the cube root of both sides, we get:
r = 3
So, when the volume is 36π cubic centimeters, the radius of the sphere is 3 centimeters.
Now, we need to find the rate of increase of the area (A) when the radius is 3 cm. To do this, we differentiate the formula for the area of a sphere with respect to time (t):
dA/dt = 4πr * dr/dt
Since dr/dt is given as 0.04 cm/s, and the radius is 3 cm, we can plug in the values:
dA/dt = 4π(3) * 0.04 cm/s
Simplifying, we find that the rate of increase of the area of a cross section through the center of the sphere is:
dA/dt ≈ 0.48π cm^2/s
c. Finally, we need to find the radius of the sphere when its volume and radius are increasing at the same numerical rate.
Let's express the rate of increase of the volume (dV/dt) and the rate of increase of the radius (dr/dt) as an equation:
dV/dt = dr/dt
Using our earlier results, we know that dV/dt = 320π cm^3/s. So we have:
320π = dr/dt
Now we can solve for the radius (r)!
Unfortunately, we don't have enough information to find the exact value of r. But hey, at least the math was fun, right?
To solve these questions, we need to use calculus and apply the concepts of rates of change and derivatives.
a. To find the rate of increase of the volume of the sphere, we need to take the derivative of the volume equation with respect to time, and then substitute the given value for the radius and the rate of change of the radius. Here's how you can do it:
1. Start with the volume equation: V = (4/3)πr^3.
2. Take the derivative of both sides of the equation with respect to time (t). Remember, since the radius is changing with time, we need to use the chain rule. Let's call the derivative of the volume (dV/dt) and the derivative of the radius (dr/dt):
dV/dt = (4/3)π * 3r^2 * (dr/dt)
Simplifying this expression gives us:
dV/dt = 4πr^2 * (dr/dt)
3. Substitute the given values into the equation. In this case, the radius (r) is 10 cm, and the rate of increase of the radius (dr/dt) is 0.04 cm/s:
dV/dt = 4π(10)^2 * 0.04
dV/dt = 4π(100) * 0.04
dV/dt = 16π cm^3/s
Therefore, at the time when the radius of the sphere is 10 cm, the rate of increase of its volume is 16π cubic centimeters per second.
b. To find the rate of increase of the area of a cross-section through the center of the sphere, we need to calculate the derivative of the area equation with respect to time. Here's what to do:
1. The area of a cross-section through the center of the sphere is A = πr^2, where r is the radius.
2. Take the derivative of both sides of the equation with respect to time (t). Again, we need to use the chain rule since the radius is changing with time. Call the derivative of the area (dA/dt) and the derivative of the radius (dr/dt):
dA/dt = π * 2r * (dr/dt)
Simplifying this expression gives us:
dA/dt = 2πr * (dr/dt)
3. We need to find the rate of increase of the area when the volume of the sphere is 36π cubic centimeters. We can use the relationship between volume and radius to determine the radius at that volume.
V = (4/3)πr^3
36π = (4/3)πr^3
Simplify the equation:
36 = (4/3)r^3
Multiply by 3/4 to isolate r^3:
r^3 = 27
Calculate the cube root of both sides:
r = 3 cm
4. Substitute the radius value into the equation for the rate of change of the area:
dA/dt = 2π(3) * (0.04)
dA/dt = 0.24π cm^2/s
Therefore, at the time when the volume of the sphere is 36π cubic centimeters, the rate of increase of the area of a cross-section through the center of the sphere is 0.24π square centimeters per second.
c. To find the radius at the time when the volume and the radius of the sphere are increasing at the same numerical rate, we need to equate the rates of change of the volume and the radius:
1. The rate of increase of the volume (dV/dt) is given by the equation from part a: dV/dt = 4πr^2 * (dr/dt).
2. The rate of increase of the radius (dr/dt) is also given as 0.04 cm/s.
3. Set the rates of change equal to each other:
4πr^2 * (dr/dt) = 0.04
4. Simplify the equation:
4πr^2 = 0.04
Divide both sides by 4π:
r^2 = 0.01
5. Take the square root of both sides to find the radius:
r = 0.1 cm
Therefore, at the time when the volume and the radius of the sphere are increasing at the same numerical rate, the radius is 0.1 cm.