Chapter 13: Problem 77
A particle moves along a path \(C\) given by the equations \(x=t\) and \(y=-t^{2}\). If \(z=x^{2}+y^{2}\), find \(d z / d s\) along \(C\) at the instant when the particle is at the point \((2,-4)\).
Short Answer
Expert verified
\( \frac{dz}{ds} = \frac{36}{\sqrt{17}} \) at the point (2,-4).
Step by step solution
01
Understanding the Problem
We are given a particle moving along a path defined by the parametric equations \( x(t) = t \) and \( y(t) = -t^2 \). We want to find \( \frac{dz}{ds} \) at a specific point on this path, where \( z = x^2 + y^2 \) and \( s \) represents the arc length.
02
Expressing \( z \) in Terms of \( t \)
Substitute the parametric equations into the expression for \( z \):\[ z(t) = [x(t)]^2 + [y(t)]^2 = t^2 + (-t^2)^2 = t^2 + t^4. \]
03
Differentiate \( z(t) \) with Respect to \( t \)
Differentiate \( z(t) = t^2 + t^4 \) with respect to \( t \): \[ \frac{dz}{dt} = \frac{d}{dt}(t^2 + t^4) = 2t + 4t^3.\]
04
Calculate the Arc Length Derivative \( \frac{ds}{dt} \)
The arc length element \( ds \) along the path is given by \( \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt. \) First find \( \frac{dx}{dt} = 1 \) and \( \frac{dy}{dt} = -2t \). Then compute \( \frac{ds}{dt} = \sqrt{1^2 + (-2t)^2} = \sqrt{1 + 4t^2}. \)
05
Calculate \( \frac{dz}{ds} \)
Using the chain rule, \( \frac{dz}{ds} = \frac{dz}{dt} \cdot \frac{dt}{ds} \). Given \( \frac{dt}{ds} = \frac{1}{\sqrt{1 + 4t^2}} \), we find: \( \frac{dz}{ds} = (2t + 4t^3) \cdot \frac{1}{\sqrt{1 + 4t^2}}. \)
06
Determine \( t \) at Point (2,-4)
The point \((2,-4)\) corresponds to \( x(t) = t = 2 \) and \( y(t) = -t^2 = -4 \). Both equations confirm \( t = 2 \).
07
Evaluate \( \frac{dz}{ds} \) at \( t = 2 \)
Substitute \( t = 2 \) into \( \frac{dz}{ds} = (2t + 4t^3) \cdot \frac{1}{\sqrt{1 + 4t^2}} \): \[ \frac{dz}{ds} \bigg|_{t=2} = (2(2) + 4(2)^3) \cdot \frac{1}{\sqrt{1 + 4(2)^2}} = (4 + 32) \cdot \frac{1}{\sqrt{17}} = \frac{36}{\sqrt{17}}.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
In mathematics, parametric equations are a powerful tool used to define a set of related quantities called parameters. Unlike traditional functions that connect one variable to another, parametric equations describe a path, curve or surface using one or more parameters, effectively establishing a position in space. For example, in the exercise, we use two parametric equations to describe a particle's motion:
- \( x(t) = t \)
- \( y(t) = -t^2 \)
Arc Length
Arc length is a crucial concept that measures the distance along a curve or path. In parametric terms, the arc length \( s \) from a point \( t = a \) to \( t = b \) can be calculated by integrating an arc length element over the parameter range. The arc length \( ds \) is given by:\[d s = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]For the particle’s path described in the exercise, we find:
- \( \frac{dx}{dt} = 1 \)
- \( \frac{dy}{dt} = -2t \)
- So, the arc length element becomes \( \sqrt{1 + 4t^2} \, dt \).
Chain Rule
The Chain Rule is one of calculus' fundamental concepts, enabling us to differentiate composite functions. It states that if you have a function \( z \) that is dependent on another variable \( y \), which in turn is dependent on \( x \), then the derivative of \( z \) with respect to \( x \) is:\[\frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx}\]In the context of the exercise, we have \( z = x^2 + y^2 \) and both \( x \) and \( y \) are functions of \( t \). We need to find \( \frac{dz}{ds} \), where \( s \) is the arc length. Thus, using the Chain Rule:
- Find \( \frac{dz}{dt} \): Already calculated as \( 2t + 4t^3 \)
- Compute \( \frac{dt}{ds} \): Obtained as \( \frac{1}{\sqrt{1 + 4t^2}} \)
- Apply the Chain Rule: \( \frac{dz}{ds} = \frac{dz}{dt} \cdot \frac{dt}{ds} \)