Problem 200 - Calculus Volume 3

Question

The centripetal acceleration of a particle moving in a circle is given by \(a(r, v)=\frac{v^{2}}{r},\) where \(v\) is the velocity and \(r\) is the radius of the circle. Approximate the maximum percent error in measuring the acceleration resulting from errors of \(3 \%\) in \(v\) and \(2 \%\) in \(r\). (Recall that the percentage error is the ratio of the amount of error over the original amount. So, in this case, the percentage error in \(a\) is given by \(\frac{d a}{a} .\) )

Step-by-Step Solution

Answer

The maximum approximated percent error in acceleration is 4%.

Step 1: Write down the formula for percentage error

To solve the problem, we need to approximate the percentage error in acceleration, \(a\), which is given by the expression \(\rac{da}{a}\)\. We use differentials to find \(rac{da}{a}\) for \(a(r,v) = \frac{v^2}{r}\)\.

Step 2: Find the differential of the acceleration

First, we calculate the differential using derivatives. The partial derivatives are \( \frac{\partial a}{\partial v} = \frac{2v}{r} \) and \( \frac{\partial a}{\partial r} = -\frac{v^2}{r^2} \)\. The total differential \(da\) is:\[ da = \left(\frac{\partial a}{\partial v}\right) dv + \left(\frac{\partial a}{\partial r}\right) dr = \frac{2v}{r}dv - \frac{v^2}{r^2}dr \]\.

Step 3: Express percentage errors in terms of differentials

The percentage errors in \(v\) and \(r\) are given as \(\frac{dv}{v} = 0.03\) and \(\frac{dr}{r} = 0.02\) respectively. Substitute these into the differential expression for \(da\):\[ da = \frac{2v}{r}(0.03v) - \frac{v^2}{r^2}(0.02r) \]\.

Step 4: Simplify the expression for fractional error

Substitute \(dv = 0.03v \) and \(dr = 0.02r \) into the total differential \(da\):\[ da = \frac{2v}{r} \times 0.03v - \frac{v^2}{r^2} \times 0.02r = \frac{0.06v^2}{r} - \frac{0.02v^2}{r} \]\. Simplifying gives: \(da = \frac{0.04v^2}{r} \)\.

Step 5: Calculate the percentage error in acceleration

Now, we find the percentage error \(\frac{da}{a}\) by dividing \(da = \frac{0.04v^2}{r}\) by the original acceleration \(a = \frac{v^2}{r}\):\[\frac{da}{a} = \frac{0.04v^2/r}{v^2/r} = 0.04\]\. This means there is a \(4\%\) error in the estimation of acceleration.

Key Concepts

Percentage Error

The concept of percentage error is crucial in understanding how measurements can deviate from expected or exact values. In simple terms, percentage error is the ratio of the error's magnitude to the actual or expected value, expressed as a percentage.

• It helps in quantifying the uncertainty in measurements.
• In the context of centripetal acceleration, we calculate the percentage error for the acceleration function.

The exercise involves errors in both the measurement of velocity (\(v\)) and the radius (\(r\)), causing an error in acceleration (\(a\)). The percentage error here is given by \(\frac{da}{a}\), where \(da\) is the differential difference in the acceleration and \(a\) is the actual acceleration.

Differentials

Differentials provide a way to approximate changes in functions based on their rates of change.
- In this exercise, differentials help us approximate the change in centripetal acceleration when the velocity or radius changes slightly.

• The differential \(da\) is expressed as a sum of changes due to \(v\) and \(r\):
• \(da = \left(\frac{\partial a}{\partial v}\right) dv + \left(\frac{\partial a}{\partial r}\right) dr\)

These partial derivatives show how sensitive the acceleration is to changes in each variable. By computing these, we can predict the impact of small measurement errors on the centripetal acceleration. This calculation results in understanding how differential calculus is applied in error analysis.

Partial Derivatives

Partial derivatives are used when functions depend on multiple variables, like our centripetal acceleration function \(a(r, v) = \frac{v^{2}}{r}\).
- They measure how a function changes as one particular variable changes, keeping others constant. This is crucial when approximating the error in acceleration due to errors in velocity and radius.
The two partial derivatives here \(\frac{\partial a}{\partial v} = \frac{2v}{r}\) and \(\frac{\partial a}{\partial r} = -\frac{v^2}{r^2}\) reflect:

• Change in acceleration with respect to velocity.
• Change in acceleration with respect to radius.

By using these, we can add up the individual impacts, assessing the full effect of these small changes, thus showing how partial derivatives enable multi-variable function analysis.

Velocity and Radius

Velocity and radius are two key parameters in the function for centripetal acceleration.
Changing these directly affects the centripetal force and, consequently, the acceleration.
- **Velocity (\(v\))**: Represents how fast an object is moving along the circular path. The square of the velocity \(v^2\) is directly proportional to acceleration. An increase in velocity results in a significant ramp-up in acceleration. For errors, a 3% change is substantial.
- **Radius (\(r\))**: Represents how large the circle is. It’s inversely proportional to acceleration; a larger radius decreases the acceleration since force spreads over a longer path. A 2% error here means radius variations have a smaller but still meaningful impact on acceleration.
When calculating centripetal acceleration errors, understanding these relationships helps in predicting how sensitive the acceleration will be to measurement inaccuracies in \(v\) and \(r\). This insight is essential in any experimental or practical scenarios involving circular motion.