You plan to paint a sphere of radius 200 cm with a coat of pain .2 cm thick. Use differentials to approximate the amount of paint that is needed. Use the equation for the volume of a sphere.
v = 4/3 πr^3
dv = 4πr^2 dr
so, dv = 4π*200^2*.2 = 32000π cm^3
note that the dv is just the surface area times the thickness, as you'd expect for tiny thicknesses.
To approximate the amount of paint needed, we can use differentials.
The equation for the volume of a sphere is given by V = (4/3)πr^3, where V is the volume and r is the radius.
Let's differentiate both sides of the equation with respect to r:
dV = d((4/3)πr^3)
Now, we can use differentials to approximate the change in volume.
dV ≈ (4/3)π(3r^2)dr
Since we are given that the radius of the sphere is 200 cm and the thickness of the paint is 0.2 cm, we can substitute these values into our equation:
dV ≈ (4/3)π(3(200)^2)(0.2) (approximation for dV)
Simplifying this expression:
dV ≈ (4/3)π(3)(40,000)(0.2)
dV ≈ (4/3)π(24,000)
dV ≈ 32,000π cm^3
So, the approximate amount of paint needed is 32,000π cm^3.
To approximate the amount of paint needed to coat the sphere, we can use differentials.
The volume of a sphere is given by the equation V = (4/3)πr³, where V is the volume and r is the radius.
In this case, the radius of the sphere is 200 cm and the thickness of the paint coat is .2 cm. We want to find the change in volume of the sphere when the radius increases by .2 cm.
First, we need to find the derivative of the volume equation with respect to the radius. Taking the derivative of V = (4/3)πr³ gives us dV/dr = 4πr².
Next, we substitute the given radius into the derivative equation: dV/dr = 4π(200)² = 4π(40000) = 160000π.
Now, multiply the derivative by the change in radius: dV = (160000π)(.2) = 32000π.
This gives us the change in volume of the sphere when the radius increases by .2 cm. However, we only need to approximate the amount of paint needed, so we can ignore the π and get an approximate answer of 32000 cubic centimeters (cc).