<item ident="F1-0688" title="F1 | Direction fields for first-order ODEs | ver. 0688">
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<p>
<strong>F1.</strong>
</p>
<p> Use <a href="https://sagecell.sagemath.org/">https://sagecell.sagemath.org/</a> to run the SageMath code <code>t,y = var('t y'); plot_slope_field(cos(y/2), (t,-5,5), (y,-5,5))</code> producing the direction field for the ODE <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y'} = \cos\left(\frac{1}{2} \, {y}\right)" alt="{y'} = \cos\left(\frac{1}{2} \, {y}\right)" title="{y'} = \cos\left(\frac{1}{2} \, {y}\right)" data-latex="{y'} = \cos\left(\frac{1}{2} \, {y}\right)"/>. </p>
<p> Let <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y_p" alt="y_p" title="y_p" data-latex="y_p"/> be the solution to the following IVP. Explain how to use its direction field to approximate the value of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y_p" alt="y_p" title="y_p" data-latex="y_p"/> at <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?t= 1" alt="t= 1" title="t= 1" data-latex="t= 1"/>. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y'} = \cos\left(\frac{1}{2} \, {y}\right) \hspace{2em} y( -1 )= 0" alt="{y'} = \cos\left(\frac{1}{2} \, {y}\right) \hspace{2em} y( -1 )= 0" title="{y'} = \cos\left(\frac{1}{2} \, {y}\right) \hspace{2em} y( -1 )= 0" data-latex="{y'} = \cos\left(\frac{1}{2} \, {y}\right) \hspace{2em} y( -1 )= 0"/>
</p>
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<p>
<strong>F1.</strong>
</p>
<p> Use <a href="https://sagecell.sagemath.org/">https://sagecell.sagemath.org/</a> to run the SageMath code <code>t,y = var('t y'); plot_slope_field(cos(y/2), (t,-5,5), (y,-5,5))</code> producing the direction field for the ODE <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By'%7D%20=%20%5Ccos%5Cleft(%5Cfrac%7B1%7D%7B2%7D%20%5C,%20%7By%7D%5Cright)" alt="{y'} = \cos\left(\frac{1}{2} \, {y}\right)" title="{y'} = \cos\left(\frac{1}{2} \, {y}\right)" data-latex="{y'} = \cos\left(\frac{1}{2} \, {y}\right)">. </p>
<p> Let <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y_p" alt="y_p" title="y_p" data-latex="y_p"> be the solution to the following IVP. Explain how to use its direction field to approximate the value of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y_p" alt="y_p" title="y_p" data-latex="y_p"> at <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?t=%201" alt="t= 1" title="t= 1" data-latex="t= 1">. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By'%7D%20=%20%5Ccos%5Cleft(%5Cfrac%7B1%7D%7B2%7D%20%5C,%20%7By%7D%5Cright)%20%5Chspace%7B2em%7D%20y(%20-1%20)=%200" alt="{y'} = \cos\left(\frac{1}{2} \, {y}\right) \hspace{2em} y( -1 )= 0" title="{y'} = \cos\left(\frac{1}{2} \, {y}\right) \hspace{2em} y( -1 )= 0" data-latex="{y'} = \cos\left(\frac{1}{2} \, {y}\right) \hspace{2em} y( -1 )= 0">
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<h4>Partial Answer:</h4>
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<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y_p( 1 )\approx 1.7" alt="y_p( 1 )\approx 1.7" title="y_p( 1 )\approx 1.7" data-latex="y_p( 1 )\approx 1.7"/>
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<h4>Partial Answer:</h4>
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<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y_p(%201%20)%5Capprox%201.7" alt="y_p( 1 )\approx 1.7" title="y_p( 1 )\approx 1.7" data-latex="y_p( 1 )\approx 1.7">
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