Chapter 10: Problem 50
Find \(d y / d x\) and \(d^{2} y / d x^{2}\) at the given point without eliminating the parameter. $$ x=\cos \phi, y=3 \sin \phi ; \phi=5 \pi / 6 $$
Short Answer
Expert verified
\( \frac{dy}{dx} = 3\sqrt{3}, \frac{d^2 y}{dx^2} = -48 \)
Step by step solution
01
Differentiate x with respect to φ
Given that \( x = \cos \phi \), we find \( \frac{dx}{d\phi} \) by differentiating:\[ \frac{dx}{d\phi} = -\sin \phi \]
02
Differentiate y with respect to φ
Given that \( y = 3\sin \phi \), we differentiate to find \( \frac{dy}{d\phi} \):\[ \frac{dy}{d\phi} = 3 \cos \phi \]
03
Use chain rule to find dy/dx
Using the chain rule formula \( \frac{dy}{dx} = \frac{dy/d\phi}{dx/d\phi} \), plug in the derivatives found:\[ \frac{dy}{dx} = \frac{3 \cos \phi}{-\sin \phi} = -3 \cot \phi \]
04
Substitute φ to find dy/dx at φ = 5π/6
Substitute \( \phi = \frac{5\pi}{6} \) into the expression \(-3 \cot \phi\):\[ \cot \frac{5\pi}{6} = -\sqrt{3} \] (since \( \sin \frac{5\pi}{6} = \frac{1}{2} \) and \( \cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} \))Thus,\[ \frac{dy}{dx} = -3(-\sqrt{3}) = 3\sqrt{3} \]
05
Find the second derivative formula d²y/dx²
To find \( \frac{d^2 y}{dx^2} \), we use the formula:\[ \frac{d^2 y}{dx^2} = \frac{d}{d\phi}\left(\frac{dy}{dx}\right) \cdot \frac{d\phi}{dx} = \frac{d}{d\phi}\left(-3 \cot \phi \right) \cdot \left(-\frac{1}{\sin \phi}\right) \]
06
Differentiate -3 cot φ and simplify expression
Differentiate \(-3 \cot \phi\):\[ \frac{d}{d\phi}(-3 \cot \phi) = 3 \csc^2 \phi \]Then, we plug in the expression to find \( \frac{d^2 y}{dx^2} \):\[ \frac{d^2 y}{dx^2} = 3 \csc^2 \phi \cdot \left(-\frac{1}{\sin \phi}\right) = -3 \frac{\csc^3 \phi}{\sin \phi} \]
07
Substitute φ to find d²y/dx² at φ = 5π/6
Substitute \( \phi = \frac{5\pi}{6} \):\[ \csc \frac{5\pi}{6} = 2 \] (since \( \sin \frac{5\pi}{6} = \frac{1}{2} \))Thus, applying to \( -3 \frac{\csc^3 \phi}{\sin \phi} \), we get:\[ -3 \cdot \frac{2^3}{1/2} = -3 \cdot \frac{8}{1/2} = -48 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First Derivative
When we talk about finding the first derivative in a parametric context, we're focusing on determining how one variable changes with respect to another as they are both defined in terms of a parameter. In the case of our example, we are given two functions: \( x = \cos \phi \) and \( y = 3\sin \phi \). Here, \( \phi \) is the parameter. To find the first derivative \( \frac{dy}{dx} \), we follow a specific process:
- First, differentiate \( x \) with respect to \( \phi \): \( \frac{dx}{d\phi} = -\sin \phi \).
- Next, differentiate \( y \) with respect to \( \phi \): \( \frac{dy}{d\phi} = 3 \cos \phi \).
- Finally, apply the chain rule to combine these results and find \( \frac{dy}{dx} = \frac{dy/d\phi}{dx/d\phi} \).
Second Derivative
Calculating the second derivative, \( \frac{d^2 y}{dx^2} \), provides insights into the concavity of the curve at a specific point, which tells us how the rate of change itself is changing. This can be important for understanding the curvature of the graph.To find \( \frac{d^2 y}{dx^2} \), use the first derivative expression found earlier. You differentiate this first derivative with respect to \( \phi \), and then multiply by \( \frac{d\phi}{dx} \), as shown in the solution steps:
- Start with the derivative of \(-3 \cot \phi\). The differentiation yields \( 3 \csc^2 \phi \).
- Then, multiply by \( \frac{d\phi}{dx} = -\frac{1}{\sin \phi} \) to account for the change in \( \phi \) with \( x \).
Chain Rule
The chain rule is a fundamental technique in calculus that allows us to differentiate composite functions. In parametric differentiation, it is especially useful because we often deal with functions of a parameter, not directly expressed in terms of each other. Therefore, the chain rule gives us a tool to relate these functions.In our example:
- We have \( y = 3\sin \phi \) and \( x = \cos \phi \), both in terms of \( \phi \).
- To find \( \frac{dy}{dx} \), which is the derivative of \( y \) with respect to \( x \), we apply the chain rule as: \( \frac{dy}{dx} = \frac{dy/d\phi}{dx/d\phi} \).