Linear Approximations with a hemispherical dome

[forestmine]

Homework Statement

Use linear approximations to estimate the amount of paint needed to apply a
coat of pain .05cm thick to a hemispherical dome with diameter 50m.

Homework Equations

V=2/3*$\pi$*r$^{3}$

The Attempt at a Solution

I'm really not at all sure how to go about this. The only thing I somewhat understand is that it's necessary to subtract the difference in volume to obtain just the volume of the shell itself.

If someone could walk me through this one, that'd be great.

Thanks!

 

[mighty2000]

Hello,

Thank you for your question. Linear approximation is a useful tool in estimating values that are difficult to calculate directly. In this case, we can use linear approximation to estimate the volume of paint needed to apply a coat of paint 0.05cm thick to a hemispherical dome with a diameter of 50m.

To begin, let's first calculate the volume of the entire hemisphere using the formula V = 2/3 * π * r^3, where r is the radius of the hemisphere. Since the diameter is given to be 50m, the radius would be half of that, which is 25m. Therefore, the volume of the hemisphere would be V = 2/3 * π * (25m)^3 = 65,449.85 m^3.

Now, we need to estimate the volume of the paint needed for a 0.05cm thick coat. To do this, we can use the linear approximation formula, which is given by ΔV = V' * Δx, where ΔV is the change in volume, V' is the derivative of the volume function with respect to x, and Δx is the change in x.

In this case, x represents the thickness of the paint and Δx represents the change in thickness, which is 0.05cm. The derivative of the volume function with respect to x would be V' = 4/3 * π * r^2.

Substituting the values, we get ΔV = (4/3 * π * (25m)^2) * (0.05cm) = 261.8 m^3.

Therefore, the estimated volume of paint needed for a 0.05cm thick coat would be V_est = V + ΔV = 65,449.85 m^3 + 261.8 m^3 = 65,711.65 m^3.

I hope this helps. Let me know if you have any further questions or if you need clarification on any of the steps. Good luck with your calculations!