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Chapter 3: Problem 11

Carry out the following steps. a. Use implicit differentiation to find \(\frac{d y}{d x}\) b. Find the slope of the curve at the given point. $$\cos y=x ;\left(0, \frac{\pi}{2}\right)$$

Short Answer

Expert verified
Answer: The slope of the curve at the given point is -1.

Step by step solution

01

Implicit differentiation of \(\cos y = x\)

To find the derivative \(\frac{dy}{dx}\), differentiate both sides of the equation with respect to \(x\): $$ \frac{d (\cos y)}{d x} = \frac{d x}{d x} $$ Now, apply the chain rule on the left side by differentiating \(\cos y\) with respect to \(y\) and multiplying it with \(\frac{dy}{dx}\) (since we are differentiating with respect to \(x\) and not \(y\)): $$ -\sin y \frac{dy}{dx} = 1 $$ Now, solve for \(\frac{dy}{dx}\): $$ \frac{dy}{dx} = -\frac{1}{\sin y} $$
02

Find the slope at the given point

Now, we need to find the slope at the given point \((0, \frac{\pi}{2})\). To do this, plug in x = 0 and y = \(\frac{\pi}{2}\) into \(\frac{dy}{dx}\): $$ \frac{dy}{dx}\Big|_{(0, \frac{\pi}{2})} = -\frac{1}{\sin (\frac{\pi}{2})} $$ Since \(\sin(\frac{\pi}{2}) = 1\), $$ \frac{dy}{dx}\Big|_{(0, \frac{\pi}{2})} = -\frac{1}{1} = -1 $$ So, the slope of the curve at the given point \((0, \frac{\pi}{2})\) is -1.

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Key Concepts

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

Chain Rule
At its core, the chain rule is a fundamental technique in calculus for differentiating composite functions. When a function is composed of one function inside of another, the chain rule allows us to find the derivative of the entire composition by taking the derivative of the outer function and multiplying it by the derivative of the inner function.

In the context of the exercise above, implicit differentiation employs the chain rule. Since we're given the equation involving a cosine function of y (\f\(\f(cos y\f) = x\f\)), we treat y as a function of x. Thus, when we differentiate \f\(\f(cos y\f)\) with respect to x, we first take the derivative of the cosine function (which is \f\(-\f(sin y)\f\)) and then multiply by the derivative of y with respect to x (\f\(\f(dy/dx)\f\)). This results in \f\(-\f(sin y) \f(dy/dx)\f\) which is the application of the chain rule in this exercise.

This rule is essential in differentiating not just explicit functions but also implicit functions where y is not isolated on one side of an equation.
Slope of a Curve
The slope of a curve at any point refers to the steepness or the incline of the curve at that specific point. It can be considered as the instantaneous rate of change of the function at a certain point. The concept of slope is fundamental in understanding the behavior of lines and curves in calculus.

In simpler terms, the slope tells us how much y changes for a small change in x around that point. For linear functions, this is constant. However, for curves like the one in our exercise, the slope can change at every point. To find the slope of the curve at a particular point, we determine the derivative of the function, which represents the slope of the tangent to the curve at any given point.

Once we have the derivative from the implicit differentiation step, as we saw in the exercise (\f\(\f(dy/dx) = -1/ \f(sin y)\f\)), we can find the slope at a specific point by substituting the x and y values of the point into this derivative expression. At the point \f\((0, \f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(pi/2\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\f)\), the slope is \f\(-1\f\). This means, the curve descends to the right of the point at a consistent rate.
Derivative Calculus
Derivative calculus is the branch of mathematics that deals with finding and interpreting the derivatives of functions. The derivative of a function at a point provides the rate at which the function changes at that point, which is analogous to the concept of velocity in physics. It's a measure of how a function's output value changes as its input changes.

The act of finding a derivative is called differentiation. For functions y that are explicitly defined as a function of x (like \f\(y = x^2\f\)), taking a derivative is often straightforward. However, for implicitly defined functions, where y is not isolated, implicit differentiation, as shown in the original exercise, comes into play.

Derivative calculus is a crucial part of the calculus curriculum because it's applied in a variety of scientific and engineering domains, from analyzing the motion of objects to optimizing functions in economics, and it forms the basis for more advanced topics in calculus such as integrals and differential equations.

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