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Magazine Chapter 11: Problem 1
In Exercises \(1-14\) , find an equation for the line tangent to the curve at the point defined by the given value of \(t\) . Also, find the value of \(d^{2} y / d x^{2}\) at this point. $$ x=2 \cos t, \quad y=2 \sin t, \quad t=\pi / 4 $$
Short Answer
Expert verified
Tangent line: \(y = -x + 2\sqrt{2}\); \(d^2y/dx^2 = 0\).
Step by step solution
01
Determine Coordinates at Given t
We need to find the coordinates of the point on the curve at the given value of \( t = \frac{\pi}{4} \).For \( x = 2 \cos t \):\[ x = 2 \cos \frac{\pi}{4} = 2 \times \frac{\sqrt{2}}{2} = \sqrt{2} \]For \( y = 2 \sin t \):\[ y = 2 \sin \frac{\pi}{4} = 2 \times \frac{\sqrt{2}}{2} = \sqrt{2} \]So, the point is \((\sqrt{2}, \sqrt{2})\).
02
Find dx/dt and dy/dt
To find the slope of the tangent line, we need to calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).\( \frac{dx}{dt} = \frac{d}{dt}(2 \cos t) = -2 \sin t \)\( \frac{dy}{dt} = \frac{d}{dt}(2 \sin t) = 2 \cos t \)
03
Calculate dy/dx at t = 蟺/4
Using the chain rule, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).Plug in the value \( t = \frac{\pi}{4} \):\[ \frac{dy}{dx} = \frac{2 \cos \frac{\pi}{4}}{-2 \sin \frac{\pi}{4}} = \frac{\sqrt{2}}{-\sqrt{2}} = -1 \]The slope of the tangent line at the point \((\sqrt{2}, \sqrt{2})\) is \(-1\).
04
Write the Equation of the Tangent Line
The equation of the tangent line in point-slope form is given by:\[ y - y_1 = m(x - x_1) \]Where \( m = -1 \) is the slope, and \((x_1, y_1) = (\sqrt{2}, \sqrt{2})\) is the point.Therefore, the equation becomes:\[ y - \sqrt{2} = -1(x - \sqrt{2}) \]Simplifying gives:\[ y = -x + 2\sqrt{2} \]
05
Calculate d虏y/dx虏
To find \( \frac{d^2y}{dx^2} \), we need the derivative of \( \frac{dy}{dx} \) with respect to \( t \) divided by \( \frac{dx}{dt} \).We have:\( \frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy}{dx}) \div \frac{dx}{dt} \)Given \( \frac{dy}{dx} = -1 \), \( \frac{d}{dt}(-1) = 0 \),thus \[ \frac{d^2y}{dx^2} = \frac{0}{-2 \sin t} = 0 \] at \( t = \frac{\pi}{4} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
Parametric equations are a set of equations that express the coordinates of the points that make up a geometric shape as functions of a parameter, usually denoted as \( t \). Instead of defining \( y \) as a direct function of \( x \), parametric equations approach the problem by defining both \( x \) and \( y \) as separate functions of \( t \). This allows for more complex curves to be defined.
- In our exercise, the parametric equations are \( x = 2 \cos t \) and \( y = 2 \sin t \). They represent a circle with radius 2 due to the trigonometric identities.
- For example, at \( t = \frac{\pi}{4} \), both \( x \) and \( y \) can be computed to give the coordinates \( (\sqrt{2}, \sqrt{2}) \) on the circle.
Differentiation
Differentiation is a core concept in calculus that deals with finding the rate at which a function is changing at any point. In the context of parametric equations:
- Differentiation helps us find the slope of the tangent to the curve, which is necessary to write its equation.
- Finding \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) lets us measure how \( x \) and \( y \) change with respect to \( t \).
Chain Rule
The chain rule is a critical concept in calculus used to differentiate composite functions. It states that the derivative of a composition of functions is the derivative of the outer function times the derivative of the inner function.
- When dealing with parametric equations, the chain rule is employed to find \( \frac{dy}{dx} \) by using the derivatives \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \).
- Specifically, in our problem, it allowed us to compute the slope \( \frac{dy}{dx} \) even when \( y \) is not a direct function of \( x \). We used the chain rule to write \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).
Curvature
Curvature is a measure of how sharply a curve bends at any given point. It gives an idea of the geometric bending, or how the direction of the tangent line is changing as \( t \) changes.
- The curvature is mathematically represented through the second derivative, \( \frac{d^2y}{dx^2} \), which gives the concavity of the curve.
- In our particular problem, we found that \( \frac{d^2y}{dx^2} = 0 \), indicating that around \( t = \frac{\pi}{4} \), the tangent line doesn't change its slope rapidly.
