Question

if the sides of a cube are measured with an error of 2% use differentials to estimate the relative error in the volume

df= 3x^2 * dx
df= 3x^2 * 1/50
df= 3/50*x^2

is this right?

Answer 1

v = x^3
dv = 3 x^2 dx agrred
dv/v *100 = percent error in volume

dv/v = 3 x^2/x^3 dx

but dx/x * 100 = percent error in side

dv/v = 3 dx/x
so
dv/v * 100 = 3 dx/x * 100

= 3 * 2% = 6 percent

Answer 2

Thank you!! :-)