Problem 17 - University Calculus: Early Transcendentals

Question

Distance Let \(x\) and \(y\) be differentiable functions of \(t\) and let \(s=\sqrt{x^{2}+y^{2}}\) be the distance between the points \((x, 0)\) and \((0, y)\) in the \(x y\) -plane. a. How is \(d s / d t\) related to \(d x / d t\) if \(y\) is constant? b. How is \(d s / d t\) related to \(d x / d t\) and \(d y / d t\) if neither \(x\) nor \(y\) is constant? c. How is \(d x / d t\) related to \(d y / d t\) if \(s\) is constant?

Step-by-Step Solution

Answer

a) Relates to \(\frac{dx}{dt}\) as \(\frac{ds}{dt} = \frac{x}{\sqrt{x^2 + y^2}} \cdot \frac{dx}{dt}\). b) Relates by \(\frac{ds}{dt} = \frac{x}{\sqrt{x^2 + y^2}} \cdot \frac{dx}{dt} + \frac{y}{\sqrt{x^2 + y^2}} \cdot \frac{dy}{dt}\). c) \(\frac{dx}{dt} = -\frac{y}{x} \cdot \frac{dy}{dt}\).

Step 1: Understanding the Derivative for Part (a)

Given that \( y \) is constant, the expression for the distance \( s = \sqrt{x^2 + y^2} \) becomes \( s = \sqrt{x^2 + c^2} \, where \ c \) is a constant. Differentiate both sides with respect to \( t \), recognizing that the derivative of a constant is zero. Apply the chain rule to obtain \( \frac{ds}{dt} = \frac{x}{\sqrt{x^2 + y^2}} \cdot \frac{dx}{dt} \), since \( \frac{dy}{dt} = 0 \).

Step 2: Applying the Chain Rule for Part (b)

In this scenario, neither \( x \) nor \( y \) is constant. Differentiate the expression \( s = \sqrt{x^2 + y^2} \) with respect to \( t \) using implicit differentiation and the chain rule. The result is \[ \frac{ds}{dt} = \frac{x}{\sqrt{x^2 + y^2}} \cdot \frac{dx}{dt} + \frac{y}{\sqrt{x^2 + y^2}} \cdot \frac{dy}{dt}. \] This equation shows how \( \frac{ds}{dt} \) relates to both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

Step 3: Analyzing Constant Distance for Part (c)

If \( s \) is constant, \( \frac{ds}{dt} = 0 \. \) From part (b), we have \( \frac{x}{\sqrt{x^2 + y^2}} \cdot \frac{dx}{dt} + \frac{y}{\sqrt{x^2 + y^2}} \cdot \frac{dy}{dt} = 0 \. \) Simplifying, we find \[ x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt} = 0. \] Solving for one variable in terms of the other gives \( \frac{dx}{dt} = -\frac{y}{x} \cdot \frac{dy}{dt} \).

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus for differentiating composite functions. It is essential when dealing with functions that are composed of two or more functions. For instance, when you have

• a function inside another function,
• or when the variable you're differentiating is also a function of another variable,

you use the chain rule to find the rate of change of that entire composition.

In the context of the exercise, consider the composite function of distance, represented as \( s = \sqrt{x^2 + y^2} \). This function depends on both \( x \) and \( y \), which are in turn dependent on \( t \). Thus, to differentiate this function with respect to \( t \), the derivative must account for the changes in both \( x \) and \( y \) as well.

By applying the chain rule, we achieve:

• For a constant \( y \), \( \frac{ds}{dt} = \frac{x}{\sqrt{x^2 + y^2}} \cdot \frac{dx}{dt} \).
• When neither \( x \) nor \( y \) is constant, \( \frac{ds}{dt} = \frac{x}{\sqrt{x^2 + y^2}} \cdot \frac{dx}{dt} + \frac{y}{\sqrt{x^2 + y^2}} \cdot \frac{dy}{dt} \).

The chain rule thus easily extends to more complex functions and makes differentiating them much more manageable.

Derivatives

Derivatives are one of the core concepts of calculus, and they represent how a function changes as its input changes. In simple terms, a derivative is the rate of change, or slope, of a function at any point on a curve.

In mathematical terms, if you have a function \( f(t) \), the derivative of this function with respect to \( t \) is denoted by \( f'(t) \) or \( \frac{df}{dt} \).

In the exercise

• the quantity \( x(t) \) and \( y(t) \) represent some form of movement or change occurring over time \( t \).
• The derivative \( \frac{dx}{dt} \) indicates how fast \( x \) is changing as time \( t \) progresses.
• Similarly, \( \frac{dy}{dt} \) reflects how fast \( y \) is changing as time \( t \) progresses.

Understanding the role of derivatives allows us to predict or analyze the behavior of dynamic processes over time. By calculating these rates of change, we get insights into how distance \( s \) changes concerning \( x \) and \( y \). Derivatives provide the mathematical framework needed for solving problems involving velocity and other real-world applications.

Distance Formula

The distance formula comes from the Pythagorean theorem and helps find the distance between two points on a plane. In a two-dimensional space, if you have the points \( (x_1, y_1) \) and \( (x_2, y_2) \), the distance \( s \) between these points is given by:

• \( s = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

In the context of this exercise,
the points are specifically \((x, 0)\) and \((0, y)\), making the formula simplify to \( s = \sqrt{x^2 + y^2} \).

The distance formula is incredibly useful in various fields, from geometry to physics, as it allows you to:

• exactly measure how far apart two points are in a coordinate system,
• develop further mathematical functions and identities revolving around space and distance.

Additionally, by differentiating this distance with respect to time \( t \), we can understand how the distance between these dynamic points changes over time. This concept becomes powerful in real-world applications, such as calculating the shortest path between moving objects.