2.00: Physical Problems for Numerical Differentiation
- Page ID
- 126393
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Lesson 1: General Engineering
Summary
To find the acceleration of a rocket requires differentiation of the velocity expression.
Modeling
A rocket is traveling vertically and expels fuel at a velocity of \(2000\ \text{m/s}\) at a consumption rate of \(2100\ \text{kg/s}\). The initial mass of the rocket is \(140,000\ \text{kg}\). If the rocket starts from rest, how can I calculate the acceleration of the rocket at \(16\ \text{s}\)?
If
\[m_{0} = \text{initial mass of rocket at}\ t = 0\ \text{(kg)}, \nonumber\]
\[q = \text{rate at which fuel is expelled}\ \text{(kg/s)}, \nonumber\]
\[u = \text{velocity at which the fuel is being expelled}\ \text{(m/s)} \nonumber\]
then since the fuel is expelled from the rocket, the mass of the rocket keeps decreasing with time. The mass of the rocket, \(m\) at any time, \(t\) is
\[m = m_{o} - qt \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.1}) \nonumber\]
The forces on the rocket at any time are found by applying Newton’s second law of motion. Then
\[\sum F = ma \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.2}) \nonumber\]
The upward force is due to the fuel expulsion, while the downward force is due to gravity
\[{uq} - mg = ma \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.3}) \nonumber\]
Using Equations \((\PageIndex{1.1})\) and \((\PageIndex{1.3})\) in Equation \((\PageIndex{1.2})\) gives
\[{uq} - \left( m_{o} - {qt} \right)g = \left( m_{o} - {qt} \right)a \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.4}) \nonumber\]
where
\[g = \text{acceleration due to gravity}\ \left(\text{m/s}^{2} \right) \nonumber\]
From Equation \((\PageIndex{1.4})\), we get
\[a = \frac{uq}{m_{o} - qt} - g \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.5a}) \nonumber\]
\[\frac{dv}{dt} = \frac{uq}{m_{o} - qt} - g\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.5b}) \nonumber\]
Integrating both sides of Equation \((\PageIndex{1.5b})\) gives
\[v = - u\log_{e}( m_{0} - qt) - gt + C \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.6}) \nonumber\]
Since the rocket starts from rest
\[v = 0\ \text{at}\ t = 0 \nonumber\]
\[0 = - u\log_{e}(m_{o}) + C \nonumber\]
\[C = u\log_{e}(m_{o}) \nonumber\]
Hence Equation \((\PageIndex{1.6})\) reduces to \[v = -u \log_{e} (m_{o} - qt) - gt + u \log_{e} (m_{o}) \nonumber\]
\[\frac{dx}{dt} = u \log_{e} \left( \frac{m_{o}}{m_{o} - qt} \right) -gt \quad \quad \quad (\PageIndex{1.7}) \nonumber\]
Using the following numerical values in Equation \((\PageIndex{1.7})\): \[u = 2000 \ \text{m/s} \nonumber\] \[m_{o} = 140,000 \ \text{kg} \nonumber\] \[q = 2100 \ \text{kg/s} \nonumber\] \[g = 9.8 \ \text{m/s}^{2} \nonumber\] \[t_{0} = 0 \ \text{s} \nonumber\] \[t_{1} = 30 \ \text{s} \nonumber\]
gives \[v = 2000 \log_{e} \left( \frac{14 \times 10^{4}}{14 \times 10^{4} - 2100t} \right) - 9.8t \quad \quad \quad (\PageIndex{1.8}) \nonumber\]
Can you numerically find the acceleration at \(t = 16\ \text{s}\)? You may say that we do not need numerical or analytical differentiation to calculate the acceleration, as Equation \((\PageIndex{1.5a})\) directly gives us the acceleration of the rocket at any time. True! We are just doing this as an exercise to illustrate numerical differentiation, and we have a true expression of acceleration readily available for comparison with the numerical results.
Questions
(1) Find the acceleration of the rocket at \(t = {16} \text{ s}\).
(2) Use different numerical differentiation techniques to find the acceleration at \(t = {16} \text{ s}\). Compare these results with the exact answer.
Lesson 2: Chemical Engineering
Summary
Estimating the rate at which area of containment is spreading in a lake requires numerical differentiation.
Modeling
At about 11:56 am on June 22, 1969, near the city of Cleveland in Ohio, an oil slick on the Cuyahoga River caught fire that burned for 24 minutes. This incident on a navigable river acted as a catalyst for Congress to pass the Clean Water Act in 1972. The Federal Water Pollution Control Act prohibits the discharge of oil or oily waste substances or hazardous substances into or upon the navigable waters of the United States.
Interestingly, recreational boating is a growing activity in many waterways of the United States. Unfortunately, fuel leakages – however small – from so many boats can lead to the formation of large oil slicks. The ideal would be for the recreational boats to use fuels that can evaporate as quickly as the fuels leak onto the surface of the water.
A new fuel for recreational boats being developed at the local university was tested at an area pond by a team of engineers. The interest is to document the environmental impact of the fuel – how quickly does the slick spread? Table \(\PageIndex{2.1}\) shows the video camera record of the radius of the wave generated by a drop of the fuel that fell into the pond. To find the rate at which the contamination spreads requires numerical differentiation.
\(\text{Time}\ (\text{s})\) | \(0\) | \(0.5\) | \(1.0\) | \(1.5\) | \(2.0\) | \(2.5\) | \(3.0\) | \(3.5\) | \(4.0\) | \(4.5\) | \(5.0\) |
---|---|---|---|---|---|---|---|---|---|---|---|
\(\text{Radius} \ (\text{m})\) | \(0\) | \(0.236\) | \(0.667\) | \(1.225\) | \(1.886\) | \(2.635\) | \(3.464\) | \(4.365\) | \(5.333\) | \(6.364\) | \(7.454\) |
Questions
(1) Compute the rate at which the radius of the drop was changing at \(t = 2\ \text{s}\) and \(t = 5 \text{s}\).
(2) Estimate the rate at which the area of the contaminant was spreading across the pond at \(t=2 \ \text{s}\) and \(t = 5 \ \text{s}\).