The radius of a spherical balloon is measured as 8 inches, with a possible error of 0.02inch.
a) Use differentials to approximate the possible propagated error in computing the volume of the sphere.
b) Use differentials to approximate the possible propagated error in computing the surface area of the sphere.
c) Approximate the percent errors in part a and b
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As for part a, I got +-5.12pi in^3
Part b) I got +-1.28pi in^3
C) part a=0.75% error
part b=0.5% error
I got the same answers.
I hope we're both right!
a) Well, isn't it interesting how even balloons like to inflate their egos by measuring their radii? Anyway, let's use differentials to approximate the possible propagated error in computing the volume of the sphere.
The volume of a sphere is given by V = (4/3)πr^3, where r is the radius. To find the propagated error, we differentiate this equation with respect to r:
dV = (4/3)π(3r^2)dr
Now, we substitute the given radius and error values: r = 8 inches and dr = 0.02 inches:
dV = (4/3)π(3(8)^2)(0.02)
Simplifying, we get:
dV ≈ (4/3)π(192)(0.02)
Calculating, we find:
dV ≈ 5.12π in^3
So, the possible propagated error in computing the volume of the sphere is approximately ±5.12π in^3. Don't worry, though – it won't burst your mathematical bubble!
b) Alright, let's now use differentials to approximate the possible propagated error in computing the surface area of the sphere. Get ready for some serious spherical calculations!
The surface area of a sphere is given by A = 4πr^2. Differentiating this equation with respect to r, we find:
dA = 4π(2r)dr
Now, substituting the radius and error values: r = 8 inches and dr = 0.02 inches:
dA = 4π(2(8))(0.02)
Simplifying a bit, we get:
dA ≈ 4π(16)(0.02)
Calculating, we find:
dA ≈ 1.28π in^2
So, the possible propagated error in computing the surface area of the sphere is approximately ±1.28π in^2. Don't worry, it's just a surface change – no major cracks on the math surface!
c) Now, let's approximate the percent errors in parts a and b. Brace yourself for some percentage calculations!
To find the percent error in part a, we divide the propagated error in volume (±5.12π in^3) by the actual volume of the sphere:
Percent error = (|±5.12π in^3| / V) * 100
Since the actual volume of a sphere with radius 8 inches is (4/3)π(8^3) in^3, let's calculate:
Percent error = (|±5.12π in^3| / [(4/3)π(8^3) in^3]) * 100
Simplifying, we get:
Percent error ≈ (5.12 / 4(8^3 / 9)) * 100
Calculating, we find:
Percent error ≈ 75%
So, the approximate percent error in part a is approximately 75%. That's quite a percentage of error – as if the balloon is playing hide-and-seek with accurate measurements!
As for part b, to find the percent error, we divide the propagated error in surface area (±1.28π in^2) by the actual surface area of the sphere:
Percent error = (|±1.28π in^2| / A) * 100
Since the actual surface area of a sphere with radius 8 inches is 4π(8^2) in^2, let's calculate:
Percent error = (|±1.28π in^2| / [4π(8^2) in^2]) * 100
Simplifying, we get:
Percent error ≈ (1.28 / 4(8^2)) * 100
Calculating, we find:
Percent error ≈ 50%
So, the approximate percent error in part b is approximately 50%. That's half the error, but still a considerable amount – like a clown with half the makeup but double the silliness!
Keep in mind that these approximate percent errors are based on the given values and can vary depending on the precision of measurements. So, remember to take them with a pinch of mathematical humor!
a) To approximate the possible propagated error in computing the volume of the sphere, we can use differentials.
The formula for the volume of a sphere is V = (4/3)πr^3, where r is the radius.
We are given that the radius, r, is measured as 8 inches with a possible error of 0.02 inches, so we can write the original radius as r = 8 inches and the error as Δr = 0.02 inches.
Using differentials, the propagated error in the volume ΔV is given by:
ΔV ≈ (∂V/∂r) * Δr
Differentiating the volume formula with respect to r, we get:
∂V/∂r = 4πr^2
Plugging in the values, we have:
∂V/∂r = 4π(8)^2 = 256π
Now we can calculate the propagated error:
ΔV ≈ 256π * 0.02
ΔV ≈ 5.12π in^3
Therefore, the possible propagated error in computing the volume of the sphere is approximately ±5.12π in^3.
b) For the surface area of the sphere, we can use a similar approach.
The formula for the surface area of a sphere is A = 4πr^2.
Using differentials, the propagated error in the surface area ΔA is given by:
ΔA ≈ (∂A/∂r) * Δr
Differentiating the area formula with respect to r, we get:
∂A/∂r = 8πr
Plugging in the values, we have:
∂A/∂r = 8π(8) = 64π
Now we can calculate the propagated error:
ΔA ≈ 64π * 0.02
ΔA ≈ 1.28π in^2
Therefore, the possible propagated error in computing the surface area of the sphere is approximately ±1.28π in^2.
c) To approximate the percent errors, we can use the following formulas:
Percent error = (Propagated error / Original value) * 100
For part a:
Percent error = (5.12π / (4/3)π(8)^3) * 100
Percent error ≈ 0.75%
For part b:
Percent error = (1.28π / 4π(8)^2) * 100
Percent error ≈ 0.5%
Therefore, the approximate percent errors for part a and part b are 0.75% and 0.5% respectively.
To solve this problem, we can use differential calculus to approximate the possible propagated errors in computing the volume and surface area of a spherical balloon.
a) To approximate the possible propagated error in computing the volume of the sphere, we can use differentials. The volume of a sphere is given by V = (4/3)πr^3, where r is the radius.
We are given that the radius, r, is measured as 8 inches with a possible error of 0.02 inches. So we have r = 8 inches and Δr = 0.02 inches.
Using differentials, we can calculate ΔV, the possible propagated error in the volume:
ΔV = dV/dr * Δr
To find dV/dr, we differentiate the volume formula:
dV/dr = d/dx[(4/3)πx^3]
= (4/3)π * 3x^2
= 4πx^2
Substituting r = 8 inches into dV/dr, we get:
dV/dr = 4π(8)^2
= 256π
Now we can calculate ΔV:
ΔV = 256π * Δr
= 256π * 0.02
= 5.12π
Therefore, the possible propagated error in the volume of the sphere is ±5.12π in^3.
b) To approximate the possible propagated error in computing the surface area of the sphere, we can use differentials again. The surface area of a sphere is given by A = 4πr^2.
Using the same values for r and Δr as in part a, we can calculate ΔA, the possible propagated error in the surface area:
ΔA = dA/dr * Δr
To find dA/dr, we differentiate the surface area formula:
dA/dr = d/dx[4πx^2]
= 8πx
Substituting r = 8 inches into dA/dr, we get:
dA/dr = 8π(8)
= 64π
Now we can calculate ΔA:
ΔA = 64π * Δr
= 64π * 0.02
= 1.28π
Therefore, the possible propagated error in the surface area of the sphere is ±1.28π in^2.
c) To approximate the percent errors, we need to calculate the ratios of the errors to the original values and multiply them by 100.
For part (a), the percent error is given by:
Percent error = (ΔV / V) * 100
= (5.12π / (4/3)π(8)^3) * 100
= 0.75%
For part (b), the percent error is given by:
Percent error = (ΔA / A) * 100
= (1.28π / 4π(8)^2) * 100
= 0.5%
Therefore, the approximated percent errors for part (a) and part (b) are 0.75% and 0.5%, respectively.