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By Daniel A. Murray

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4 _4. 4 _5. Slope of pA for the interval 6-8 is v(6) = ___ pA(8) = pA(6) + v(6) * 2 = ____ 24 _6. 2 _7. 8) = 16 Lesson 11 Name: Graphing Solutions to Differential Equations Date: There is another force, called drag, that acts on a falling apple, and on any object moving through air. At low velocities drag is very small, but it increases rapidly as velocity increases. One formula for calculating drag is Fdrag = Cdrag * dair * v2 * cs / 2 where Fdrag is the force due to drag Cdrag is the drag coefficient for the shape of the object dair is the density of air in kg/m3 v2 is v*v where v is the objects velocity in m/sec cs is the cross-sectional area of the object in m2 Yes, we have the numbers.

Any two points on the graph can be used to draw the straight line. Lesson 1b Graphing Solutions to Differential Equations EXERCISES _1. Draw a linear graph that has slope 10 and goes through the point (5, 50) 170 Points on the graph: (0,20) (1,___) (2, 30) . (10, ___) 100 Points on the graph: (0. ___) . (5,50) (6, 53) . ___) . (10, ___) _2. Draw a linear graph that has slope 5 and goes through the point (0, 20) _3 Draw a linear graph that has slope 3 and goes through the point (5, 50) 95 _4. Draw a linear graph that has slope –4 and goes through the point (5, 50) Points on the graph: (0.

Write the equation for p(t) such that the graph of p has slope -5 and goes through the point (5, 50). Draw the graph below: p(t) 150 ≡ ____ + ____ * (t – ___) _6. Write the equation for p(t) such that that the graph of p has slope 3 and goes through the point (9, 60). Draw the graph below: p(t) 172 168 ≡ ____ + ____ * (t – ___) p(0) = ____ p(0) = ____ p(10) = ____ p(10) = ____ Lesson 3 Name: Graphing Solutions to Differential Equations If the graph of a linear function p has slope = V and passes through the point (a, b).

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A First Course in Infinitesimal Calculus by Daniel A. Murray


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