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Use implicit differentiation to show that \(x^{2}+y^{2}=r^{2}\) is a solution to the differential equation \(d y / d x=-x / y.\)

Short Answer

Expert verified
\(x^2 + y^2 = r^2\) is a solution because differentiating implicitly yields \(\frac{dy}{dx} = -\frac{x}{y}\).

Step by step solution

01

Differentiate Implicitly

Differentiate both sides of the equation \(x^2 + y^2 = r^2\). With respect to \(x\), the derivative is \(2x + 2y \frac{dy}{dx} = 0\).
02

Solve for \(\frac{dy}{dx}\)

Rearrange the equation \(2x + 2y \frac{dy}{dx} = 0\) to solve for \(\frac{dy}{dx}\). This gives us \(2y \frac{dy}{dx} = -2x\). Divide both sides by \(2y\) to find \(\frac{dy}{dx} = \frac{-x}{y}\).
03

Confirm Solution

The result \(\frac{dy}{dx} = \frac{-x}{y}\) is identical to the differential equation given in the problem statement: \(\frac{dy}{dx} = -\frac{x}{y}\). This confirms that \(x^2 + y^2 = r^2\) is indeed a solution to the differential equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations are equations involving derivatives of a function. They play a crucial role in describing various phenomena, such as growth, decay, oscillation, and change, in fields as diverse as physics, engineering, economics, and more. In our specific case, the differential equation given is \( \frac{dy}{dx} = -\frac{x}{y} \). This equation relates the derivative of \( y \) with respect to \( x \) to the variables \( x \) and \( y \). Differential equations can usually be classified into various types:
  • Ordinary Differential Equations (ODEs): These involve functions of one independent variable and their derivatives.
  • Partial Differential Equations (PDEs): These involve multiple independent variables.
In our example, we are dealing with an ordinary differential equation because there is only one independent variable, \( x \).To show that a certain equation is a solution to a differential equation, we differentiate it and check if the derived expression satisfies the given differential equation.
Implicit Function Theorem
The Implicit Function Theorem is a fundamental concept in calculus that provides a way to differentiate equations which are not easily solved for one variable in terms of another. For instance, the equation \( x^2 + y^2 = r^2 \) implicitly defines \( y \) as a function of \( x \), though we don't actually solve separately for \( y \).The crux of the implicit function theorem is that under certain conditions, if a function is zero at some point, you can implicitly define one of the variables as a smooth function of the others. In simpler terms, when dealing with equations where one variable cannot be easily isolated, this theorem gives us the green light to proceed with differentiation. Here's how it helps us:
  • Allows differentiation without isolating \( y \).
  • Enables calculation of \( \frac{dy}{dx} \), even when it's complex to express \( y \) explicitly.
By applying this theorem to our equation \( x^2 + y^2 = r^2 \), we can compute the derivative directly without rearranging the entire equation to express \( y \) as a standalone function of \( x \).
Derivative Computation
When computing derivatives, the rules of differentiation help us find how one variable changes with respect to another. In implicit differentiation, we differentiate each term of an equation with respect to \( x \) while treating \( y \) as an implicit function of \( x \). In our example, we are differentiating the equation \(x^2 + y^2 = r^2\):
  • For \( x^2 \), the derivative with respect to \( x \) is \( 2x \).
  • For \( y^2 \), since \( y \) is treated as a function of \( x \), we apply the chain rule, giving us \( 2y \frac{dy}{dx} \).
  • For the constant \( r^2 \), the derivative is 0, as constants have no change.
The result is an equation \( 2x + 2y \frac{dy}{dx} = 0 \). We then solve for \( \frac{dy}{dx} \), the derivative of \( y \) in terms of \( x \), to find \( \frac{dy}{dx} = \frac{-x}{y} \). This matches our original differential equation, confirming that the implicit differentiation was correctly executed.

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Most popular questions from this chapter

In the 1930 s, the Soviet ecologist G. F. Gause \(^{22}\) studied the population growth of yeast. Fit a logistic curve, \(d P / d t=k P(1-P / L),\) to his data below using the method outlined below. $$\begin{array}{l|c|c|c|c|c|c|c}\hline \text { Time (hours) } & 0 & 10 & 18 & 23 & 34 & 42 & 47 \\ \hline \text { Yeast pop } & 0.37 & 8.87 & 10.66 & 12.50 & 13.27 & 12.87 & 12.70 \\\\\hline\end{array}$$(a) Plot the data and use it to estimate (by eye) the carrying capacity, \(L\) (b) Use the first two pieces of data in the table and your value for \(L\) to estimate \(k\) (c) On the same axes as the data points, use your values for \(k\) and \(L\) to sketch the solution curve \(P=\frac{L}{1+A e^{-k t}} \quad\) where \(\quad A=\frac{L-P_{0}}{P_{0}}\)

Decide whether the statement is true or false. Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=2 x-y .\) Justify your answer. The graph of \(f\) is decreasing whenever it lies above the line \(y=2 x\) and is increasing whenever it lies below the line \(y=2 x\)

A model for the population, \(P,\) of carp in a landlocked lake at time \(t\) is given by the differential equation $$\frac{d P}{d t}=0.25 P(1-0.0004 P)$$ (a) What is the long-term equilibrium population of carp in the lake? (b) A census taken ten years ago found there were 1000 carp in the lake. Estimate the current population. (c) Under a plan to join the lake to a nearby river, the fish will be able to leave the lake. A net loss of \(10 \%\) of the carp each year is predicted, but the patterns of birth and death are not expected to change. Revise the differential equation to take this into account. Use the revised differential equation to predict the future development of the carp population.

Give an example of: A differential equation for any quantity which grows in two ways simultaneously: on its own at a rate proportional to the cube root of the amount present and from an external contribution at a constant rate.

Warfarin is a drug used as an anticoagulant. After administration of the drug is stopped, the quantity remaining in a patient's body decreases at a rate proportional to the quantity remaining. The half-life of warfarin in the body is 37 hours. (a) Sketch the quantity, \(Q,\) of warfarin in a patient's body as a function of the time, \(t,\) since stopping administration of the drug. Mark the 37 hours on your graph. (b) Write a differential equation satisfied by \(Q\) (c) How many days does it take for the drug level in the body to be reduced to \(25 \%\) of the original level?

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