# Linear approximation

• May 1st 2006, 08:59 AM
Linear approximation
Q: Use linear approximation to estimate the amount of paint in cubic centimeters needed to apply a coat of paint 0.090000 cm thick to a hemispherical dome with a diameter of 60.000 meters.

My solution:

Area of hemispherical dome = 2(pi)(d/2)^2; d = 60; looking for the volume of the coat that is 0.09cm thick on top of the area.

Therefore, I have the equation f(x) = 2(pi)(60/2)^2 * x = 1800(pi)(x).

f'(x) = x
a = 0.1
f(a) = 565.2
f'(a) = 0.1

L(a) = 0.1 ( x - 0.1) + 565.2
L(0.09) = 565.199

Answer is wrong, and I can tell something is wrong when I get f'(x).

Any hints?
• May 1st 2006, 02:51 PM
ThePerfectHacker
Quote:

Q: Use linear approximation to estimate the amount of paint in cubic centimeters needed to apply a coat of paint 0.090000 cm thick to a hemispherical dome with a diameter of 60.000 meters.

My solution:

Area of hemispherical dome = 2(pi)(d/2)^2; d = 60; looking for the volume of the coat that is 0.09cm thick on top of the area.

Therefore, I have the equation f(x) = 2(pi)(60/2)^2 * x = 1800(pi)(x).

f'(x) = x
a = 0.1
f(a) = 565.2
f'(a) = 0.1

L(a) = 0.1 ( x - 0.1) + 565.2
L(0.09) = 565.199

Answer is wrong, and I can tell something is wrong when I get f'(x).

Any hints?

The derivative of,
$f(x)=1800\pi x$ is,
$f'(x)=1800 \pi$ not $f'(x)=x$