Problem 1 - Thomas Calculus

Question

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. $$ x=2 \cos t, \quad y=2 \sin t, \quad t=\pi / 4 $$

Step-by-Step Solution

Answer

Equation: \( y = -x + 2\sqrt{2} \); \( \frac{d^2y}{dx^2} = -1 \).

Step 1: Define Parametric Derivatives

Given the parametric equations, we first find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). These will be used to find the slope of the tangent line.For \( x = 2 \cos(t) \), \( \frac{dx}{dt} = -2 \sin(t) \).For \( y = 2 \sin(t) \), \( \frac{dy}{dt} = 2 \cos(t) \).

Step 2: Calculate Slope of the Tangent Line

Use the derivatives found. The slope of the tangent line is given by \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \).Substituting the derivatives, we get:\[\frac{dy}{dx} = \frac{2 \cos(t)}{-2 \sin(t)} = -\cot(t)\]At \( t = \frac{\pi}{4} \), \( \cot\left(\frac{\pi}{4}\right) = 1 \), so the slope is \(-1\).

Step 3: Determine Point on the Curve

Substitute \( t = \frac{\pi}{4} \) into the original parametric equations to find the coordinates of the point.\[x = 2 \cos\left(\frac{\pi}{4}\right) = \sqrt{2}\]\[y = 2 \sin\left(\frac{\pi}{4}\right) = \sqrt{2}\]So, the point is \( (\sqrt{2}, \sqrt{2}) \).

Step 4: Write the Equation of the Tangent Line

With the point \((\sqrt{2}, \sqrt{2})\) and the slope \(-1\), use the point-slope form of a line equation:\[y - \sqrt{2} = -1(x - \sqrt{2})\]Simplifying, we get:\[y = -x + 2\sqrt{2}\]

Step 5: Calculate Second Derivative \( \frac{d^2y}{dx^2} \)

First, find \( \frac{d^2y}{dt^2} \) and \( \frac{d^2x}{dt^2} \).For \( \frac{d^2y}{dt^2} \), differentiate \( \frac{dy}{dt} = 2\cos(t) \) to get\[\frac{d^2y}{dt^2} = -2\sin(t)\]For \( \frac{d^2x}{dt^2} \), differentiate \( \frac{dx}{dt} = -2\sin(t) \) to get\[\frac{d^2x}{dt^2} = -2\cos(t)\]Now we use these to find \( \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left( \frac{dy}{dx} \right)}{\frac{dx}{dt}} \).Differentiating \( \frac{dy}{dx} = -\cot(t) \), gives \( \frac{d}{dt}(-\cot(t)) = \csc^2(t) \).Substituting, finally:\[\frac{d^2y}{dx^2} = \frac{\csc^2(t)}{-2\sin(t)} = -\frac{1}{2}\sin(t)\cos(t)\]At \( t = \frac{\pi}{4} \), \( \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \),so,\[\frac{d^2y}{dx^2} = -(2) \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = -1\]

Step 6: Short Answer

The tangent line is \( y = -x + 2\sqrt{2} \) and \( \frac{d^2y}{dx^2} = -1 \) at the given point.

Key Concepts

Parametric Derivatives

In calculus, parametric equations enable the expression of complex curves by defining both the x and y coordinates as functions of a third parameter, often denoted as \( t \). When working with parametric equations like \( x = 2 \cos t \) and \( y = 2 \sin t \), finding the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) is essential for determining the slope of a tangent line to the curve at any point.
Here's how you can find these derivatives:

• For \( x = 2 \cos(t) \), differentiate with respect to \( t \) to get \( \frac{dx}{dt} = -2 \sin(t) \).
• Similarly, for \( y = 2 \sin(t) \), differentiate to find \( \frac{dy}{dt} = 2 \cos(t) \).

To get the slope of the tangent line with these derivatives, you use the expression \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). This result gives the slope of the tangent at any specific value of \( t \). For this problem, at \( t = \frac{\pi}{4} \), this slope becomes \( -1 \).

Second Derivative

The second derivative in calculus helps us understand the concavity of a curve, which tells us how the slope of the tangent line is changing. In parametric form, calculating \( \frac{d^{2}y}{d x^{2}} \) is slightly more complex than in the standard Cartesian form because you need to account for changes in both \( x \) and \( y \) as functions of \( t \).
To compute the second derivative for the parametric equations given:

• Find \( \frac{d^{2}y}{dt^{2}} \), which is the derivative of \( \frac{dy}{dt} = 2\cos(t) \), resulting in \( -2\sin(t) \).
• Find \( \frac{d^{2}x}{dt^{2}} \), which is the derivative of \( \frac{dx}{dt} = -2\sin(t) \), resulting in \( -2\cos(t) \).
• The second derivative \( \frac{d^{2}y}{dx^{2}} \) then follows from substituting \( \frac{d}{dt}(-\cot(t)) = \csc^2(t) \) into \( \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} \).

Evaluating at \( t = \frac{\pi}{4} \), you substitute the known trigonometric values to determine that \( \frac{d^{2}y}{dx^{2}} = -1 \).

Point-Slope Form

The point-slope form is a method used in algebra to define the equation of a line given a point and a slope. It is expressed as \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) represents a point on the line and \( m \) is the slope.
For tangent lines, which are lines that touch a curve at a single point, this is particularly useful because you can easily incorporate the slope derived from parametric equations.
In our exercise, having the point \( (\sqrt{2}, \sqrt{2}) \) and slope \( -1 \), you plug these into the point-slope form:

• Start with \( y - \sqrt{2} = -1(x - \sqrt{2}) \).
• Simplify to get the slope-intercept form, \( y = -x + 2\sqrt{2} \).

This form is extremely handy because it gives you a straightforward equation of the tangent line, showing where it touches the curve at the specified value of \( t \).

Calculus

Calculus is an essential branch of mathematics that deals with continuous change. One of its major concepts is differentiation, which involves finding derivatives.
Derivatives help in understanding how a quantity changes as one of the variables changes, and these concepts are crucial when working with curves and shapes described in terms of functions.
In parametric equations, calculus allows us to determine:

• The first derivative \( \frac{dy}{dx} \), which gives the slope of the tangent line at any point on the curve.
• The second derivative \( \frac{d^{2}y}{d x^{2}} \), which tells us about the curve's concavity at that point.

With applications ranging from physics to engineering, calculus allows the modeling of dynamic systems and provides insights into how systems evolve over time. Understanding these concepts helps solve complex practical problems such as motion tracing, optimal design, and even predicting future trends.