Asked by Megan
A sphere is initially measured as having a radius of 4 inches. A subsequent measurement shows that the radius is really 3.99. Find (a) the actual difference in volume between the initial measurement and the correction, and (b) the difference as estimated through differentials. Note- the formula for the volume of a sphere is V=4/3(pi)r^3
I did question A, but I need help on question B...
thanks :)
I did question A, but I need help on question B...
thanks :)
Answers
Answered by
Damon
V = (4/3) pi r^3
dV/dr = (4/3)3 pi r^2 = 4 pi r^2
(which is the surface area of the sphere of course)
so
dV = 4 pi r^2 dr
here dr = - .01
so dV = -.01 * 4 * pi * 16
dV = -2.01 in^3
dV/dr = (4/3)3 pi r^2 = 4 pi r^2
(which is the surface area of the sphere of course)
so
dV = 4 pi r^2 dr
here dr = - .01
so dV = -.01 * 4 * pi * 16
dV = -2.01 in^3
Answered by
Megan
Thank you soo muccchh
Answered by
Megan
sorry actually i got that part which is part a. What I need is part (b). Find the delta(v)
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