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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 9 - Differential EquationsProblem 9.3.44b
Chapter 9, Problem 9.3.44b

brOrthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection. A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. Use the following steps to find the orthogonal trajectories of the family of ellipses 2x² + y² = a²


b. The family of trajectories orthogonal to 2x² + y² = a² satisfies the differential equation dy/dx = y/(2x). Why?

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1
Start with the given family of ellipses: \(2x^{2} + y^{2} = a^{2}\). Since \(a\) is a parameter, differentiate both sides implicitly with respect to \(x\) to find the slope \(\frac{dy}{dx}\) of the tangent to the ellipses.
Differentiating implicitly, apply the chain rule: \(\frac{d}{dx}(2x^{2}) + \frac{d}{dx}(y^{2}) = \frac{d}{dx}(a^{2})\). Since \(a\) is constant with respect to \(x\), its derivative is zero.
Calculate each derivative: \(4x + 2y \frac{dy}{dx} = 0\). Solve this equation for \(\frac{dy}{dx}\) to find the slope of the tangent lines to the ellipses.
Rearranging, we get \(2y \frac{dy}{dx} = -4x\), so \(\frac{dy}{dx} = -\frac{4x}{2y} = -\frac{2x}{y}\). This is the slope of the tangent to the original family of curves.
Orthogonal trajectories have tangent lines perpendicular to these, so their slopes satisfy \(m_{1} \cdot m_{2} = -1\). If \(m_{1} = -\frac{2x}{y}\), then the slope of the orthogonal trajectories is \(m_{2} = -\frac{1}{m_{1}} = \frac{y}{2x}\). Hence, the orthogonal trajectories satisfy \(\frac{dy}{dx} = \frac{y}{2x}\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Orthogonal Trajectories

Orthogonal trajectories are families of curves that intersect another family at right angles. At each point of intersection, the tangent lines to the curves are perpendicular, meaning their slopes multiply to -1. Finding orthogonal trajectories involves determining a new family of curves whose slopes are negative reciprocals of the original family’s slopes.

Implicit Differentiation

Implicit differentiation is used to find the derivative dy/dx when y is defined implicitly by an equation involving both x and y. By differentiating both sides of the equation with respect to x and applying the chain rule, we can express dy/dx in terms of x and y, which is essential for analyzing the slope of the given family of curves.
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Slope Relationship for Orthogonal Curves

If two curves are orthogonal, their slopes at the point of intersection satisfy the relation m1 * m2 = -1, where m1 and m2 are the slopes of the tangent lines. Given the slope of the original family, the slope of the orthogonal trajectories is the negative reciprocal. This relationship explains why the differential equation for the orthogonal trajectories is derived by taking the negative reciprocal of the original slope.
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Related Practice
Textbook Question

33–36. {Use of Tech} Computing Euler approximations Use a calculator or computer program to carry out the following steps.

b. Using the exact solution (also given), find the error in the approximation to y(T) (only at the right endpoint of the time interval).


y′(t) = -2y, y(0) = 1; Δt = 0.2, T = 2; y(t) = e⁻²ᵗ

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Textbook Question

27–30. Predator-prey models Consider the following pairs of differential equations that model a predator-prey system with populations x and y. In each case, carry out the following steps.

b. Find the lines along which x'(t) = 0. Find the lines along which y'(t) = 0.


x′(t) = 2x − 4xy, y′(t) = −y + 2xy

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Textbook Question

17–20. Increasing and decreasing solutions Consider the following differential equations. A detailed direction field is not needed.


b. In what regions are solutions increasing? Decreasing?


y'(t) = (y−1)(1+y)

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Textbook Question

Properties of stirred tank solutions


b. Verify that M(0) = M₀

Textbook Question

{Use of Tech} Tumor growth The Gompertz growth equation is often used to model the growth of tumors. Let M(t) be the mass of a tumor at time t≥0. The relevant initial value problem is 

dM/dt=−rM ln(M/K),M(0)=M0, 

where r and K are positive constants and 0<M0<K.

b. Solve the initial value problem and graph the solution for r=1,K=4, and M0=1. Describe the growth pattern of the tumor. Is the growth unbounded? If not, what is the limiting size of the tumor? 

Textbook Question

29–32. {Use of Tech} Errors in Euler’s method Consider the following initial value problems.


b. Using the exact solution given, compute the errors in the Euler approximations at t=0.2 and t=0.4.


y′(t) = 2t + 1, y(0) = 0; y(t) = t² + t

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