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Magazine Chapter 10: Problem 3
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 \(d^{2} v / d x^{2}\) at this point. $$x=4 \sin t, \quad y=2 \cos t, \quad t=\pi / 4$$
Short Answer
Expert verified
Equation of tangent line: \( y = -\frac{1}{2}x + 2\sqrt{2} \); \( \frac{d^2y}{dx^2} = -\frac{1}{4} \) at \( t = \pi/4 \).
Step by step solution
01
Find the Coordinates at t = 蟺/4
To find the coordinates of the point where the tangent line touches the curve, substitute \( t = \pi/4 \) into the given parametric equations.For \( x \), we have:\[ x = 4 \sin\left(\frac{\pi}{4}\right) = 4 \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2} \] For \( y \), we have:\[ y = 2 \cos\left(\frac{\pi}{4}\right) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2} \] Thus, the coordinates are \( (2\sqrt{2}, \sqrt{2}) \).
02
Find dx/dt and dy/dt
Differentiate the parametric equations with respect to \( t \) to find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).\[ \frac{dx}{dt} = \frac{d}{dt}(4 \sin t) = 4 \cos t \] \[ \frac{dy}{dt} = \frac{d}{dt}(2 \cos t) = -2 \sin t \]
03
Compute the Slope of the Tangent Line
The slope \( m \) of the tangent line is the derivative of \( y \) with respect to \( x \), which is \( \frac{dy/dt}{dx/dt} \).Substitute \( t = \pi/4 \):\[ m = \frac{-2 \sin(\pi/4)}{4 \cos(\pi/4)} = \frac{-2 \cdot \frac{\sqrt{2}}{2}}{4 \cdot \frac{\sqrt{2}}{2}} = -\frac{1}{2} \]
04
Write the Equation of the Tangent Line
Using the point-slope form of a line equation \( y - y_1 = m(x - x_1) \), where \( m = -1/2 \) and \( (x_1, y_1) = (2\sqrt{2}, \sqrt{2}) \), write the equation of the tangent line.\[ y - \sqrt{2} = -\frac{1}{2}(x - 2\sqrt{2}) \]
05
Simplify the Equation of the Tangent Line
Simplify the equation:\[ y = -\frac{1}{2}x + \sqrt{2} + \sqrt{2} = -\frac{1}{2}x + 2\sqrt{2} \]
06
Find the Second Derivative d虏y/dx虏
To calculate \( \frac{d^2y}{dx^2} \), start by differentiating \( \frac{dy}{dx} \) with respect to \( t \) and then dividing by \( \frac{dx}{dt} \).First, differentiate \( \frac{dy}{dx} = \frac{-2 \sin t}{4 \cos t} \):\[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(-\frac{1}{2} \tan t\right) = -\frac{1}{2} \sec^2 t \] Now find \( \frac{d^2y}{dx^2} \):\[ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{-\frac{1}{2} \sec^2(\pi/4)}{4 \cos(\pi/4)} = -\frac{1}{2} \cdot \frac{2}{4} = -\frac{1}{4} \]
07
Evaluate the Second Derivative at t = 蟺/4
Substitute \( t = \pi/4 \) into the expression for \( \frac{d^2y}{dx^2} \) to find its value at this point, which is \( -1/4 \), confirming our simplified second derivative.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Tangent Line
A tangent line to a curve at a given point is a straight line that just "touches" the curve at that point without crossing it. Imagine a curve as the path of a moving object. At any point, the tangent line represents the direction in which the object is moving at that precise instant. To find the equation of a tangent line, use the point-slope form
- First, identify the point on the curve where you want the tangent.
- Next, calculate the slope of the tangent. In parametric forms, this involves taking derivatives.
- Finally, substitute these values into the point-slope form: \[ y - y_1 = m(x - x_1) \] where \( m \) is the slope at the given point \((x_1, y_1))\).
Parametric Equations
Parametric equations involve expressing the coordinates of the points on a curve as functions of a parameter, usually denoted as \( t \). Unlike other forms of equations, parametric equations can describe curves where \( y \) isn't just a direct function of \( x \). For instance,
- The "x" coordinate for a point might be \( x = 4 \sin(t) \).
- The "y" coordinate could be \( y = 2 \cos(t) \).
- Substitute the specific parameter value to find particular points on the curve.
Derivative
In calculus, the derivative is a measure of how a function changes as its input changes. It's like a "slope" or "rate of change" of the function at a particular point. When we differentiate a function, we're finding its derivative. For parametric equations, we often differentiate with respect to the parameter
- Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) for the parametric components.
- Compute the slope of the tangent line as \( \frac{dy/dt}{dx/dt} \).
Second Derivative
The second derivative of a function is essentially the derivative of the derivative of that function. It measures how the rate of change of a quantity itself is changing. With parametric equations, we derive the second derivative by differentiating the first derivative with respect to the parameter.
- Initially differentiate \( \frac{dy}{dx} \) concerning \( t \).
- Then, divide by \( \frac{dx}{dt} \) to find \( \frac{d^2y}{dx^2} \).
