A sphere's radius expands from 1.0m to 1.1m. Using differentials, what is the change in the volume of the sphere (in m3)? The volume of a sphere is given by V=4/3∏r3.
V = (4/3)π r^3
dV/dr = 4π r^2
dV = 4π r^2 dr , but r = 1 , dr = .1
= 4π(1)(.1) = .4π
Well, let's start by finding the initial volume of the sphere.
V = (4/3)πr^3
V = (4/3)π(1.0)^3
V = (4/3)π(1.0)
V = 4.18879...
Now, let's find the new volume by plugging in the expanded radius.
V' = (4/3)π(1.1)^3
V' = (4/3)π(1.331)
V' = 5.577....
To find the change in volume, we subtract the initial volume from the new volume:
ΔV = V' - V
ΔV = 5.577 - 4.18879...
ΔV ≈ 1.388...
So, the change in volume, using differentials, is approximately 1.388 cubic meters. Now that's a real growth spurt for the sphere!
To find the change in volume of a sphere using differentials, we'll differentiate the volume formula with respect to the radius.
Given: V = (4/3)πr^3
Differentiating both sides with respect to r, we get:
dV/dr = 4πr^2
Now, we need to find the change in volume (dV) when the radius changes from 1.0m to 1.1m. Since we already have the differential equation, we can substitute the radius (r) values into the equation:
dV = (4πr^2)dr
Let's substitute the initial radius (r = 1.0m):
dV = (4π(1.0^2))dr
= 4πdr
Now, let's substitute the final radius (r = 1.1m):
ΔV = 4π(1.1) - 4π(1.0)
= 4π(0.1)
= 0.4π m^3
Therefore, the change in volume of the sphere is approximately 0.4π m^3.
To find the change in volume of the sphere, we can use differentials.
Given:
Initial radius, r₁ = 1.0 m
Final radius, r₂ = 1.1 m
The volume of a sphere is given by the formula: V = (4/3)πr³
To find the change in volume, we need to find the differential of the volume with respect to the radius, dV/dr. This can be written as:
dV = (dV/dr) * dr
To differentiate V with respect to r, we can use the power rule:
dV/dr = (4/3)π * 3r²
= 4πr²
Now, we have the differential equation as:
dV = 4πr² * dr
Substituting the values of r₁ and r₂, we can find the change in volume, ΔV:
ΔV = ∫(r₁ to r₂) 4πr² * dr
Integrating this equation, we get:
ΔV = [4/3πr³] (from r₁ to r₂)
= (4/3π(1.1)³) - (4/3π(1.0)³)
= (4π/3)(1.1³ - 1.0³)
Evaluating the expression, we find the change in volume of the sphere from 1.0 m to 1.1 m is approximately:
ΔV ≈ (4π/3)(1.331 - 1.000)
≈ (4π/3)(0.331)
≈ 1.383 m³
Therefore, the change in volume of the sphere is approximately 1.383 m³.