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Chapter 3: Problem 2

Explain how implicit differentiation can simplify the work in a related-rates problem.

Short Answer

Expert verified
Answer: Implicit differentiation simplifies the work in solving related-rates problems by allowing us to take derivatives without the need to solve for the dependent variable explicitly. It focuses on the rates of change, enabling us to determine relationships between quantities and their rates of change directly from the given equation without getting caught up in the complicated relationships between the variables.

Step by step solution

01

1. Introduction to related-rates problems

Related-rates problems involve finding the rate at which a quantity changes based on its relationship to other quantities whose rates of change are known. These problems typically involve using derivatives and are often found in physics and other applied mathematics fields.
02

2. Implicit differentiation

Implicit differentiation is a technique for finding the derivative in cases when it is difficult or impossible to solve for the dependent variable explicitly. It involves taking the derivative of both sides of an equation with respect to the independent variable, while treating the dependent variable as implicitly a function of the independent variable.
03

3. Example of a related-rates problem

Suppose we have a right triangle, where the lengths of the legs are x and y, and the length of the hypotenuse is z. The relationship between the three sides is given by the Pythagorean theorem: x^2 + y^2 = z^2. Let's say we know the rates at which x and y are changing (dx/dt and dy/dt), and we want to find the rate at which z is changing (dz/dt).
04

4. Applying implicit differentiation

To find dz/dt, we first take the derivative of the Pythagorean equation with respect to time (t) using implicit differentiation: \dfrac{d}{dt}(x^2 + y^2) = \dfrac{d}{dt}(z^2) Now, we differentiate each term with respect to t, remembering that x, y, and z are all functions of t: 2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt} = 2z \dfrac{dz}{dt}
05

5. Solving for dz/dt

Now, we can solve for dz/dt: \dfrac{dz}{dt} = \dfrac{2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt}}{2z} Notice that we did not have to explicitly solve for z, which could be challenging for more complicated relationships. Implicit differentiation allows us to focus on the rates of change (dx/dt, dy/dt, and dz/dt) without getting caught up in the details of the relationship between x, y, and z.
06

6. Conclusion

Implicit differentiation simplifies the work in related-rates problems by enabling us to take derivatives without the need to solve for the dependent variable explicitly. This can save time and effort while still providing us with the information needed to tackle these types of problems.

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Use any method to evaluate the derivative of the following functions. $$f(x)=\frac{4-x^{2}}{x-2}$$

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