Maximum Error Using Differentials

In summary, the maximum error in the calculated area of the parallelogram is approximately ±5 ft^2. The differential equation used to calculate the error is correct, but the value used for Δθ should be a real number rather than in degrees.
  • #1
njo
20
0

Homework Statement


2 adjacent sides of a parallelogram measure 15ft and 10ft w/ max errors of ±0.1ft
angle is 45° w/ max error of ±0.5°

What is the maximum error in the calculated value of the area or the parallelogram?

Homework Equations


A = area = xysinθ

The Attempt at a Solution


x = 15ft dx = dy = ±0.1ft
y = 10ft dθ = ±0.5°
θ = 45°

dA = (∂A/∂x)dx + (∂A/∂y)dy + (∂A/∂θ)dθ
dA = 10sin45°(±0.1) + 15sin45°(±0.1) + 150cos45°(±0.5°)
dA = ±56.3207 ft^2

Not sure if this is correct. Tried to follow examples on my notes. Any help I would greatly appreciate. Thanks
 
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  • #2
njo said:

Homework Statement


2 adjacent sides of a parallelogram measure 15ft and 10ft w/ max errors of ±.01ft
angle is 45° w/ max error of ±0.5°

What is the maximum error in the calculated value of the area or the parallelogram?

Homework Equations


A = area = xysinθ

The Attempt at a Solution


x = 15ft dx = dy = ±0.1ft
y = 10ft dθ = ±0.5°
θ = 45°

dA = (∂A/∂x)dx + (∂A/∂y)dy + (∂A/∂θ)dθ
dA = 10sin45°(±0.1) + 15sin45°(±0.1) + 150cos45°(±0.5°)

The error in the measurements of the length of the sides is ±0.01 feet, not ±0.1 feet.

When accounting for the tolerances of angular measure, you should always use the equivalent radian measure rather than degrees.

dA = ±56.3207 ft^2

The area of the parallelogram is 106.07 ft2. An error of 56.32 ft2 would render this calculation essentially meaningless.
Not sure if this is correct. Tried to follow examples on my notes. Any help I would greatly appreciate. Thanks

You've got some changes to make.
 
  • #3
What changes? the differential equation is wrong right?
 
  • #4
njo said:
What changes? the differential equation is wrong right?
No, the differential equation appears to be correct. You just plugged the wrong numbers into it, as I explained.
 
  • #5
Sorry I didn't see your whole post on my phone at first. That's a typo. All should be +- 0.1 ft not .01 ft
 
  • #6
njo said:

Homework Statement


2 adjacent sides of a parallelogram measure 15ft and 10ft w/ max errors of ±0.1ft
angle is 45° w/ max error of ±0.5°

What is the maximum error in the calculated value of the area or the parallelogram?

Homework Equations


A = area = xysinθ

The Attempt at a Solution


x = 15ft dx = dy = ±0.1ft
y = 10ft dθ = ±0.5°
θ = 45°

dA = (∂A/∂x)dx + (∂A/∂y)dy + (∂A/∂θ)dθ
dA = 10sin45°(±0.1) + 15sin45°(±0.1) + 150cos45°(±0.5°)
dA = ±56.3207 ft^2

Not sure if this is correct. Tried to follow examples on my notes. Any help I would greatly appreciate. Thanks

Not correct, even when the ±0.1 is the correct figure.

If the two sides have lengths A and B and the angle is θ, the smallest/largest values of A are A1=14.9 and A2 =15.1, the smallest/largest values of B are B1=9.9 and B2 = 10.1, while the smallest/largest values of θ are θ1= 44.5° and θ2 = 45.5°. The smallest/largest values of the area are A1*B1*sin(θ1) and A2*B2*sin(θ2).

Compute these directly, and see how the results compare with your differential approximation. For example, you can compute Δsin(θ) = sin(45.5°)-sin(45°) exactly as written, and compare that with your calculus-based expression.
 
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  • #7
A = area = xysinθ
dA = (∂A/∂x)dx + (∂A/∂y)dy + (∂A/∂θ)dθ

So the margin for the area is about 5ft^2. How is my differential equation incorrect? I took each partial and multiplied by the rate of change.
 
  • #8
njo said:
A = area = xysinθ
dA = (∂A/∂x)dx + (∂A/∂y)dy + (∂A/∂θ)dθ

So the margin for the area is about 5ft^2. How is my differential equation incorrect? I took each partial and multiplied by the rate of change.
Your total differential is correct, but the value you used for ##\Delta \theta## is not correct. It should be a real number, not be in degrees.
 
  • #9
Mark44 said:
Your total differential is correct, but the value you used for ##\Delta \theta## is not correct. It should be a real number, not be in degrees.

That makes sense. Although I get an answer less than A2*B2*sin(θ2) - A1*B1*sin(θ1)

Δθ = .5pi/180 ≅ .0087 ∴ 10sin45°(±0.1) + 15sin45°(±0.1) + 150cos45°(±.pi/360) ≅ ±2.6934
 
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  • #10
njo said:
A = area = xysinθ
dA = (∂A/∂x)dx + (∂A/∂y)dy + (∂A/∂θ)dθ

So the margin for the area is about 5ft^2. How is my differential equation incorrect? I took each partial and multiplied by the rate of change.

You should be able to figure out this for yourself, given all the hints you received. Have you actually done what I suggested, that is, compute Δsin(θ) = sin(45.5°)-sin(45°) and then compared that with you calculus-based result?
 
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1. What is "Maximum Error Using Differentials"?

"Maximum Error Using Differentials" is a method used in calculus to estimate the maximum possible error in a function or equation. It is particularly useful when dealing with real-world data or measurements that may have some degree of uncertainty.

2. How is "Maximum Error Using Differentials" calculated?

The maximum error using differentials is calculated by taking the absolute value of the derivative of the function, multiplying it by the change in the independent variable, and then adding this value to the original function value. The resulting value is the maximum possible error in the function.

3. When should "Maximum Error Using Differentials" be used?

This method should be used when there is a need to estimate the maximum possible error in a function or equation. It is commonly used in engineering, physics, and other scientific fields where precise measurements and calculations are necessary.

4. What are the limitations of "Maximum Error Using Differentials"?

One limitation of this method is that it can only provide an estimate of the maximum error, and not an exact value. It also assumes that the error is evenly distributed throughout the function, which may not always be the case. Additionally, it is only applicable for small changes in the independent variable.

5. Can "Maximum Error Using Differentials" be applied to any type of function?

Yes, this method can be applied to any differentiable function. However, it is most commonly used for linear and polynomial functions, as well as logarithmic and exponential functions.

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