The radius of a 10 inch right circular cylinder is measured to be 4 inches, but with a possible error of ±0.1 inch. use linear approximation or differentials to determine the possible error in the volume of the cylinder. include units in your answer.

502.4 ± 30.14 in^3

Step-by-step explanation:

r = 4 in, h = 10 in

error = ± 0.1 inch

Volume of a cylinder, V = π r² h

Take log on both the sides

log V = log π + 2 log r + log h

Differentiate both sides

dV/V = 0 + 2 dr/r + dh /h

dV/V = 2 (± 0.1) / 4 + (± 0.1) / 10

dV/V = ± 0.05 ± 0.01 = ± 0.06 (1)

Now, V = 3.14 x 4 x 4 x 10 = 502.4 in^3

Put in equation (1)

dV = ± 0.06 x 502.4 = ± 30.144

So, V ± dV = 502.4 ± 30.14 in^3

30

step-by-step explanation:

times by times

x= 3^y

step-by-step explanation: