Chapter 1: Problem 84
For the parametric curve whose equation is \(x=4 \cos \theta, y=4 \sin \theta, \quad\) find the slope and concavity of the curve at \(\theta=\frac{\pi}{4}\).
Short Answer
Expert verified
Slope is -1, concavity is positive at \(\theta = \frac{\pi}{4}\).
Step by step solution
01
Find the derivative dx/dθ
To determine the slope and concavity, we first need to compute the derivatives with respect to \(\theta\). Starting with \(x = 4 \cos \theta\), differentiate it with respect to \(\theta\):\[\frac{dx}{d\theta} = -4 \sin \theta.\]
02
Find the derivative dy/dθ
Now, find the derivative of \(y\) with respect to \(\theta\). Given \(y = 4 \sin \theta\), differentiate it:\[\frac{dy}{d\theta} = 4 \cos \theta.\]
03
Calculate the slope dy/dx
The slope \(\frac{dy}{dx}\) at any point on the curve is found by using the chain rule: \(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\). Substitute the derivatives calculated above:\[\frac{dy}{dx} = \frac{4 \cos \theta}{-4 \sin \theta} = -\cot \theta.\]
04
Evaluate slope at θ=π/4
Substitute \(\theta = \frac{\pi}{4}\) into the expression for the slope:\[\frac{dy}{dx}\bigg|_{\theta=\frac{\pi}{4}} = -\cot \left(\frac{\pi}{4}\right) = -1.\]
05
Determine the second derivative d²y/dx²
To find concavity, calculate the second derivative. Start by finding \(\frac{d^2y}{d\theta^2}\) and \(\frac{d^2x}{d\theta^2}\):\[\frac{d^2x}{d\theta^2} = -4 \cos \theta, \quad \frac{d^2y}{d\theta^2} = -4 \sin \theta.\]Use these to find \(\frac{d^2y}{dx^2}\):\[\frac{d^2y}{dx^2} = \frac{\frac{d}{d\theta}(\frac{dy}{d\theta})}{\frac{dx}{d\theta}} = \frac{-4 \sin \theta}{-4 \sin \theta} = 1.\]
06
Evaluate concavity at θ=π/4
Evaluate \(\frac{d^2y}{dx^2}\) at \(\theta = \frac{\pi}{4}\):\[\frac{d^2y}{dx^2}\bigg|_{\theta=\frac{\pi}{4}} = 1.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Slope of a Parametric Curve
When you have a parametric curve, both the x and y coordinates are defined in terms of a third parameter, often called \( \theta \). This means the curve is a set of points \((x(\theta), y(\theta))\). To find the slope of such a curve, you calculate \( \frac{dy}{dx} \), which is the rate of change of \( y \) with respect to \( x \). This tells you how steep the curve is at any given point.
To find \( \frac{dy}{dx} \) for a parametric curve, use the chain rule from calculus. First, find \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \). These represent the rate of change for y and x concerning \( \theta \). Then, the slope \( \frac{dy}{dx} \) is simply \( \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \).
In our example, for \( x = 4 \cos \theta \) and \( y = 4 \sin \theta \), the derivatives are \( \frac{dx}{d\theta} = -4 \sin \theta \) and \( \frac{dy}{d\theta} = 4 \cos \theta \). Thus, \( \frac{dy}{dx} = -\cot \theta \), showing the slope at any point is defined by the cotangent of \( \theta \).
To find \( \frac{dy}{dx} \) for a parametric curve, use the chain rule from calculus. First, find \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \). These represent the rate of change for y and x concerning \( \theta \). Then, the slope \( \frac{dy}{dx} \) is simply \( \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \).
In our example, for \( x = 4 \cos \theta \) and \( y = 4 \sin \theta \), the derivatives are \( \frac{dx}{d\theta} = -4 \sin \theta \) and \( \frac{dy}{d\theta} = 4 \cos \theta \). Thus, \( \frac{dy}{dx} = -\cot \theta \), showing the slope at any point is defined by the cotangent of \( \theta \).
- It gives insight into the direction and steepness of the curve.
- If the slope is positive, the curve rises to the right.
- A negative slope means it falls to the right.
Concavity of a Curve
The concavity of a curve indicates whether the curve is curving upwards (concave up) or downwards (concave down) at a particular point. It is determined by the second derivative \( \frac{d^2y}{dx^2} \).
For parametric equations, finding this second derivative involves a few steps: Compute \( \frac{d^2x}{d\theta^2} \) and \( \frac{d^2y}{d\theta^2} \). The formula for \( \frac{d^2y}{dx^2} \) then becomes \( \frac{\frac{d}{d\theta}(\frac{dy}{d\theta})}{\frac{dx}{d\theta}} \).
In the provided problem, we find \( \frac{d^2x}{d\theta^2} = -4 \cos \theta \) and \( \frac{d^2y}{d\theta^2} = -4 \sin \theta \). Using these, the second derivative is \( 1 \), meaning the curve is concave up at \( \theta = \frac{\pi}{4} \).
For parametric equations, finding this second derivative involves a few steps: Compute \( \frac{d^2x}{d\theta^2} \) and \( \frac{d^2y}{d\theta^2} \). The formula for \( \frac{d^2y}{dx^2} \) then becomes \( \frac{\frac{d}{d\theta}(\frac{dy}{d\theta})}{\frac{dx}{d\theta}} \).
In the provided problem, we find \( \frac{d^2x}{d\theta^2} = -4 \cos \theta \) and \( \frac{d^2y}{d\theta^2} = -4 \sin \theta \). Using these, the second derivative is \( 1 \), meaning the curve is concave up at \( \theta = \frac{\pi}{4} \).
- A positive second derivative indicates concave up behavior.
- A negative second derivative indicates concave down behavior.
Calculus Derivatives
Derivatives are key in calculus for understanding how functions change. They provide a powerful way to analyze and describe motion and change, especially with curves. For parametric curves, derivatives help express changes concerning the parameter, usually \( \theta \).
To facilitate the understanding:
To facilitate the understanding:
- First derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) describe the velocity of each coordinate.
- They describe how the x and y positions change as \( \theta \) changes.
- The second derivatives \( \frac{d^2x}{d\theta^2} \) and \( \frac{d^2y}{d\theta^2} \) describe acceleration.
- Concavity, from the second derivative \( \frac{d^2y}{dx^2} \), tells whether a curve is bending upwards or downwards.