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Differential Equations - Applying Differential Equations

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    Study Guide Differential Equations – Applying Differential Equations 1. Applications of First ‐ Order Equations First - order differential equations show up in many real - world problems — from geometry and physics to chemistry and biology. In this chapter, we explore several important applications and learn how to model and solve them. 1.1 Orthogonal Trajectories What does “orthogonal” mean? • Orthogonal means perpendicular . • A trajectory is a path or curve. So, orthogonal trajectories are two families of curves that intersect each other at right angles (90°). 1.2 How do we know two curves are perpendicular? At any point where two curves intersect: • Their slopes must be negative reciprocals of 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 curves are orthogonal wherever they intersect if: Example 1: Electric Field Lines and Equipotentials

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    Study Guide Figure 1 Electric field lines from a positive point charge spread outward in straight lines from the charge. Step 1: Describe the electric field lines If 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 family Differentiate: From ( y = cx ) , we get: So the slope becomes:

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    Study Guide Step 3: Find the orthogonal trajectories Orthogonal trajectories must have slopes that are negative reciprocals : This equation is separable. Step 4: Solve the differential equation Integrate both sides: Rewriting: Figure 2 Final Result The orthogonal trajectories are circles centered at the origin .

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    Study Guide Interpretation: • 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 Circles Figure 3 We are given the family: These are circles tangent to the x - axis at the origin . Step 1: Differentiate implicitly Expand and differentiate:

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    Study Guide Differentiate: Solve for (dy/dx ) : Step 2: Eliminate the parameter ( c ) From the original equation: Substitute into the slope expression: Step 3: Find the orthogonal trajectories Take the negative reciprocal: This leads to the differential equation: Step 4: Solve using an integrating factor The equation is not exact, but an integrating factor ( \ mu = x^{ - 2} ) makes it exact.

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    Study Guide After integrating: Figure 4 Final Result The orthogonal trajectories are also circles tangent to the x - axis at the origin . 1.3 Radioactive Decay Some atomic nuclei are unstable and decay over time. The rate of decay depends on how much of the substance is present. The basic model Let ( x(t) ) be the amount of radioactive material at time ( t ) : Solution of the equation This separable equation gives: This curve is called an exponential decay curve .

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    Study Guide Half - Life The half - life ( T 1/2 ) is the time it takes for half of the substance to decay. From the formula: A shorter half - life means faster decay. 1.4 Radiocarbon Dating • Carbon - 14 ( 14 C) is radioactive. • Living organisms constantly absorb it. • After death, absorption stops and decay begins. Since the half - life of ( 14 C ) is 5730 years , measuring how much remains allows scientists to estimate age. Example 3: Dating a Bone A bone contains 20% of its original ( 14 C). Using: Solve for ( t ) : 1.5 Newton’s Law of Cooling When a hot object is placed in a cooler environment, its temperature decreases over time.

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    Study Guide The law states: where: • ( T ) = object temperature • ( T s ) = surrounding temperature • ( k > 0 ) Solution Solving gives: Example 4: Cooling Coffee (Newton’s Law of Cooling) A cup of coffee at 190°F is placed in a room where the temperature is 70°F . After 5 minutes , the coffee cools to 160°F . Question: How many more minutes will it take for the coffee to cool to 130°F ? Step 1: Use Newton’s Law of Cooling Newton’s Law of Cooling says that the rate at which an object cools is proportional to the difference between its temperature and the surrounding temperature. The temperature model is:

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    Study Guide where: • ( T s = 70°F ) (room temperature) • ( c ) and ( k ) are constants • ( t ) is time in minutes So 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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    Study Guide Solve step by step: Step 4: Write the final temperature formula Substitute ( k ) into the equation: Step 5: Find when the coffee reaches 130°F Set ( T(t) = 130 ) : Solve: Final Answer It takes 12 minutes total for the coffee to cool from 190°F to 130°F. Since the coffee already cooled for 5 minutes , the additional time required is:

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