Answer:
96 cc
102.125 cc
90.125 cc
Step-by-step explanation:
Let V(x) the volume of a cube of edge x cm. Then
[tex] \bf V(x)=x^3 \;cm^3[/tex]
The maximum possible error in computing the volume of the cube would be
V(8+0.5) - V(8)
By approximating with the derivative, we know
V(8+0.5) ≅ V(8) +0.5V'(8)
hence
V(8+0.5) - V(8) ≅ 0.5V'(8)
But
[tex] \bf V'(x)=3x^2\Rightarrow V'(8)=3*64=192 [/tex]
and maximum possible error in computing the volume of the cube would be approximately 0.5*192 = 96 cc.
find the actual error in measuring volume if the radius was really 8.5 cm instead of 8 cm
[tex] \bf V(8.5) - V(8) = (8.5)^3-8^3=614.125-512=102.125\;cm^3[/tex]
find the actual error if the radius was actually 7.5 cm instead of 8 cm.
[tex] \bf V(8) - V(7.5) = (8)^3-(7.5)^3=512-421.875=90.125\;cm^3[/tex]
