2: Write an equation for the line in slope-intercept form.
2 answers:
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It's A. B doesn't really make sense ("derivative of the differential equation" is somewhat nonsensical, "derivative of an equation" is not meaningful).
More to the point: Slope fields are used to visualize solutions to differential equations of the form
<em>y'</em> = <em>f(x</em>, <em>y)</em>
You take some point (<em>x</em>, <em>y</em>) and evaluate <em>y'</em> at the point. This gives the slope of the line tangent to the particular solution to the DE that passes through the point (<em>x</em>, <em>y </em>).
Sample several points and evaluate <em>y'</em> at those points and you get several different slopes.
Simple example:
<em>y'</em> = <em>x</em> ² - <em>y</em> ² = (<em>x</em> + <em>y </em>) (<em>x</em> - <em>y</em> )
Let's take the points (1, 1), (-1, 0), and (2, -2), at which we get slopes
<em>y'</em> = <em>f</em> (1, 1) = 0
<em>y'</em> = <em>f</em> (-1, 0) = 1
<em>y'</em> = <em>f</em> (2, -2) = 0
From here, you can get particular solutions that pass through a certain point by interpolating the slopes of the tangents to the solution. I've attached a slope field for the example here at the points listed above. (In the order red, green, black). Each light gray arrow in the background shows the slope of the tangent line.
If you're still unsure how the slope field is generated, I suggest looking up videos on the subject. The process is a bit difficult to describe without a dynamic visual aid.
