• asd1249jf
In summary, the problem is calculating the absolute error in the result of using the derivative with respect to z.
asd1249jf
Hello.

I've encountered a problem when trying to attack this problem:

A right triangle has sides of length x, y and z. Measurements show that legs x and y have lengths of 119ft and 120ft, respectively. The percentage relative uncertainty in each measurement is 1%. For x, this means that 100 * dx / x = 1. (That 1 would be reported as 1%)

A)Use the pythagorean theorem to calculate the percentage relative uncertainty in z

B)Use the law of cosines and your result from part a to calculate the percentage relative uncertainty in the right angle.

So for part A, z = (x^2 + y^2)^.5

therefore,

dz = (.5(2x)^-.5)dx + (.5(2y)^-.5)dy

if I simply substitute the values in for x, dx, y and dy I get a bizarre answer for the percentage relative unceratainty, and 100*dz/z should be 1. Where am I going wrong with this problem?

and for part B)

z^2 = x^2 + y^2 - 2xycos(theta)

so the partial derivation would be

2z = -2xy(-sin(theta) * curlyd(theta)/curlydz)

but I am completely stuck where to go from here.

A help will be greatly appriciated!

l46kok said:
Hello.

I've encountered a problem when trying to attack this problem:

A right triangle has sides of length x, y and z. Measurements show that legs x and y have lengths of 119ft and 120ft, respectively. The percentage relative uncertainty in each measurement is 1%. For x, this means that 100 * dx / x = 1. (That 1 would be reported as 1%)

A)Use the pythagorean theorem to calculate the percentage relative uncertainty in z

B)Use the law of cosines and your result from part a to calculate the percentage relative uncertainty in the right angle.

So for part A, z = (x^2 + y^2)^.5

therefore,

dz = (.5(2x)^-.5)dx + (.5(2y)^-.5)dy

if I simply substitute the values in for x, dx, y and dy I get a bizarre answer for the percentage relative unceratainty, and 100*dz/z should be 1. Where am I going wrong with this problem?
Why should 100*dz/z be 1? That's what you are asked to find. Also notice that dx and dy are the actual "uncertainties", not the "percentage relative" but you are given "percentage relative" uncertainties when you are told 1%. That is what is meant by 100 dx/x= 1. dx= 0.1(x) and dy= 0.1(y). Finally, you are differentiating wrong there should be a factor of $(x^2+ y^2)^{-0.5}$. It's simpler to use implicit differentiation: if $x^2+ y^2= z^2$, then $2x= 2z\frac{\partial z}{\partial x}$ so that $\frac{\partial z}{\partial x}= \frac{x}{z}$.
Of course, $\frac{\partial z}{\partial y}= \frac{y}{z}$ so that
$$dz= \frac{x}{z}dx+ \frac{y}{z}dy$$.

and for part B)

z^2 = x^2 + y^2 - 2xycos(theta)

so the partial derivation would be

2z = -2xy(-sin(theta) * curlyd(theta)/curlydz)

but I am completely stuck where to go from here.

A help will be greatly appriciated!
Why are you using only the derivative with respect to z?
$0= 2x- 2ycos(\theta)+ 2xy sin(\theta)\frac{\partial \theta}{\partial x}$ so
$$\frac{\partial \theta}{\partial x}= \frac{2x- 2y cos(\theta)}{2xy sin(\theta)}$$
$$\frac{\partial \theta}{\partial y}= \frac{2y- 2x cos(\theta)}{2xy sin(\theta)}$$
$$\frac{\partial \theta}{\partial z}= \frac{2z}{-2xy sin(\theta)}$$

Then
$$d\theta= \frac{\partial \theta}{\partial x}dx+ \frac{\partial \theta}{\partial y}dy+ \frac{\partial \theta}{\partial z}dz$$

Vo =x√(g/2h)
can you please help me ...i have to find the absolute error using the derivate .Does anyone knows the formula?

1. What is partial derivation and how is it used to calculate percentage relative uncertainty?

Partial derivation is a mathematical technique used to calculate the change in a dependent variable based on small changes in one or more independent variables. In the context of calculating percentage relative uncertainty, partial derivation allows us to determine how the uncertainty in one variable affects the overall uncertainty in a final result.

2. Why is it important to calculate percentage relative uncertainty?

Percentage relative uncertainty allows us to quantify the amount of uncertainty in a measurement or calculation. This is crucial in scientific research, as it helps us understand the reliability and accuracy of our data and results.

3. What are the steps involved in using partial derivation to calculate percentage relative uncertainty?

The first step is to determine the formula for the final result, which includes one or more independent variables. Then, we calculate the partial derivatives of that formula with respect to each independent variable. Finally, we plug in the values for each variable and their respective uncertainties to calculate the overall percentage relative uncertainty.

4. Can you give an example of using partial derivation to calculate percentage relative uncertainty?

Sure, let's say we have a formula for calculating the volume of a cube: V = s^3, where s is the length of one side. If the measured length of the side is 10 cm with an uncertainty of 0.1 cm, the percentage relative uncertainty in the volume can be calculated as: (3s^2)(0.1 cm) / (s^3) * 100% = 0.3%.

5. Are there any limitations or assumptions when using partial derivation to calculate percentage relative uncertainty?

Yes, partial derivation assumes that the uncertainties in each independent variable are independent of each other. It also assumes that the uncertainties are small enough to be approximated by the first-order derivative. Additionally, partial derivation may not account for all sources of uncertainty in a measurement or calculation, so it should be used with caution and in conjunction with other methods of uncertainty analysis.

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