- #1
buffgilville
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A hemispherical dome with radius 48 feet will be given a coat of paint 0.07 inches thick. Use differentials to estimate the number of gallons of paint that will be needed to paint the dome.
Here's what I did:
f(x+h) = f '(x)h + f(x) and v=4/3pi*r^3; r=48ft.
volume of the dome:
V=4/3pi*r^3 = f(r)
I set (r+h) to be the radius of the paint volume
so, V=4/3pi*(r+h)^3 = f(r+h)
then, f(r+h) = f '(r)h + f(r)
The question just want the volume of paint needed, so I subtracted f(r)
f(r+h)=f '(r)h + f(r) - f(r), simplify
then, f(r+h)=f'(r)h
derivative, V(paint) = 4/3pi*r^3, then dv/dr(paint)=4pi*r^2
given: r=48 feet, h=0.07inches = 0.0058333feet
f(r+h)=f'(r)h => f(r+h) = 4pi*(48)^2*0.0058333
I got approximately 168.891056 gallons, but the correct answer is 631.7000268 gallons.
What did I go wrong?
Here's what I did:
f(x+h) = f '(x)h + f(x) and v=4/3pi*r^3; r=48ft.
volume of the dome:
V=4/3pi*r^3 = f(r)
I set (r+h) to be the radius of the paint volume
so, V=4/3pi*(r+h)^3 = f(r+h)
then, f(r+h) = f '(r)h + f(r)
The question just want the volume of paint needed, so I subtracted f(r)
f(r+h)=f '(r)h + f(r) - f(r), simplify
then, f(r+h)=f'(r)h
derivative, V(paint) = 4/3pi*r^3, then dv/dr(paint)=4pi*r^2
given: r=48 feet, h=0.07inches = 0.0058333feet
f(r+h)=f'(r)h => f(r+h) = 4pi*(48)^2*0.0058333
I got approximately 168.891056 gallons, but the correct answer is 631.7000268 gallons.
What did I go wrong?