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Differential Equations - Applying Differential Equations - Document preview page 1

Differential Equations - Applying Differential Equations - Page 1

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Differential Equations - Applying Differential Equations

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Differential Equations - Applying Differential Equations - Page 1 preview imageStudy GuideDifferential EquationsApplying Differential Equations1. Applications of FirstOrder EquationsFirst-order differential equations show up in many real-world problemsfrom geometry and physics tochemistry and biology. In this chapter, we explore several important applications and learn how tomodel and solve them.1.1Orthogonal TrajectoriesWhat does “orthogonal” mean?Orthogonalmeansperpendicular.Atrajectoryis a path or curve.So,orthogonal trajectoriesare two families of curves that intersect each other at right angles (90°).1.2How do we know two curves are perpendicular?At any point where two curves intersect:Their slopes must benegative reciprocalsof each other.If one slope is(m), the other must be(-1/m).Since the slope of a curve is given by its derivative, two families of curvesare orthogonal wherever they intersect if:Example 1:Electric Field Lines and Equipotentials
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Differential Equations - Applying Differential Equations - Page 2 preview imageStudy GuideFigure 1Electric field lines from apositive point chargespread outward in straight lines from the charge.Step 1: Describe the electric field linesIf the charge is placed at the origin, the electric field lines form the family:Different values of(c)give different straight lines through the origin.Step 2: Find the slope of the familyDifferentiate:From(y = cx), we get:So the slope becomes:
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Differential Equations - Applying Differential Equations - Page 3 preview imageStudy GuideStep 3: Find the orthogonal trajectoriesOrthogonal trajectories must have slopes that arenegative reciprocals:This equation is separable.Step 4: Solve the differential equationIntegrate both sides:Rewriting:Figure 2Final ResultThe orthogonal trajectories arecircles centered at the origin.
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Differential Equations - Applying Differential Equations - Page 4 preview imageStudy GuideInterpretation:Electric field lines are straight rays.Equipotential lines are concentric circles.These always intersect at right angles.Example 2:Orthogonal Trajectories of a Family of CirclesFigure 3We are given the family:These are circles tangent to thex-axis at the origin.Step 1: Differentiate implicitlyExpand and differentiate:
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Differential Equations - Applying Differential Equations - Page 5 preview imageStudy GuideDifferentiate:Solve for (dy/dx):Step 2: Eliminate the parameter(c)From the original equation:Substitute into the slope expression:Step 3: Find the orthogonal trajectoriesTake the negative reciprocal:This leads to the differential equation:Step 4: Solve using an integrating factorThe equation is not exact, but an integrating factor(\mu = x^{-2})makes it exact.
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Differential Equations - Applying Differential Equations - Page 6 preview imageStudy GuideAfter integrating:Figure 4Final ResultThe orthogonal trajectories arealso circles tangent to the x-axis at the origin.1.3Radioactive DecaySome atomic nuclei are unstable and decay over time. Therate of decaydepends on how much ofthe substance is present.The basic modelLet(x(t))be the amount of radioactive material at time(t):Solution of the equationThis separable equation gives:This curve is called anexponential decay curve.
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Differential Equations - Applying Differential Equations - Page 7 preview imageStudy GuideHalf-LifeThehalf-life(T1/2)is the time it takes for half of the substance to decay.From the formula:A shorter half-life means faster decay.1.4Radiocarbon DatingCarbon-14 (14C) is radioactive.Living organisms constantly absorb it.After death, absorption stops and decay begins.Since the half-life of (14C) is5730 years, measuring how much remains allows scientists to estimateage.Example 3:Dating a BoneA bone contains20%of its original (14C).Using:Solve for(t):1.5Newton’s Law of CoolingWhen a hot object is placed in a cooler environment, its temperature decreases over time.
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Differential Equations - Applying Differential Equations - Page 8 preview imageStudy GuideThe law states:where:(T)= object temperature(Ts)= surrounding temperature(k > 0)SolutionSolving gives:Example 4:Cooling Coffee (Newton’s Law of Cooling)A cup of coffee at190°Fis placed in a room where the temperature is70°F.After5 minutes, the coffee cools to160°F.Question:How manymore minuteswill it take for the coffee to cool to130°F?Step 1: Use Newton’s Law of CoolingNewton’s Law of Cooling says that the rate at which an object cools is proportional to the differencebetween its temperature and the surrounding temperature.The temperature model is:
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Differential Equations - Applying Differential Equations - Page 9 preview imageStudy Guidewhere:(Ts= 70°F)(room temperature)(c)and(k)are constants(t)is time in minutesSo we start with:Step 2: Find the constant(c)We know the initial temperature is 190°F:Substitute:Now the temperature model becomes:Step 3: Find the cooling constant(k)We are told that after 5 minutes, the temperature is 160°F:
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Differential Equations - Applying Differential Equations - Page 10 preview imageStudy GuideSolve step by step:Step 4: Write the final temperature formulaSubstitute(k)into the equation:Step 5: Find when the coffee reaches 130°FSet(T(t) = 130):Solve:Final AnswerIt takes12 minutes totalfor the coffee to cool from 190°F to 130°F.Since the coffee already cooled for5 minutes, theadditional time requiredis:
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