Problem 10 - 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. $$ x=1 / t, \quad y=-2+\ln t, \quad t=1 $$

Step-by-Step Solution

Answer

The tangent line equation is \( y = -x - 1 \), and \( \frac{d^2y}{dx^2} = 1 \) at this point.

Step 1: Find the Point on the Curve

First, we need to find the coordinates of the point on the curve corresponding to the given parameter value \( t = 1 \). Use the equations for \( x \) and \( y \):\( x = \frac{1}{t} = \frac{1}{1} = 1 \), and \( y = -2 + \ln(t) = -2 + \ln(1) = -2 \). Thus, the point is \((1, -2)\).

Step 2: Differentiate to Find dx/dt and dy/dt

Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) by differentiating the parametric equations:\( \frac{dx}{dt} = -\frac{1}{t^2} \) \( \frac{dy}{dt} = \frac{1}{t} \).

Step 3: Find the Slope of the Tangent Line (dy/dx)

Using \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \), substitute the derivatives to find the slope:\[ \frac{dy}{dx} = \frac{\frac{1}{t}}{-\frac{1}{t^2}} = -t \].Substitute \( t = 1 \) to get the slope: \( \frac{dy}{dx} = -1 \).

Step 4: Write the Equation of the Tangent Line

With the slope from the previous step and point \((1, -2)\), use the point-slope form, \( y - y_1 = m(x - x_1) \), to find the equation:\[ y + 2 = -1(x - 1) \]\[ y = -x - 1 \].

Step 5: Find the Second Derivative d^2y/dx^2

Find \( \frac{d^2y}{dx^2} \) using the chain rule:Given \( \frac{dy}{dx} = -t \), differentiate with respect to \( t \):\( \frac{d}{dt}\left(-t\right) = -1 \).Then, using the formula \( \frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy}{dx}) \div \frac{dx}{dt} \):\[ \frac{d^2y}{dx^2} = \frac{-1}{-\frac{1}{t^2}} = t^2 \].Substitute \( t = 1 \) to get \( \frac{d^2y}{dx^2} = 1 \).

Key Concepts

Understanding Parametric Equations

Parametric equations provide a convenient method to represent curves by defining both the x and y coordinates in terms of a third variable, commonly called "parameter". In mathematical contexts, this parameter is usually denoted by the letter \( t \). Instead of expressing y explicitly as a function of x, parametric equations express both x and y as functions of \( t \). For example, in the original exercise, the equations \( x = \frac{1}{t} \) and \( y = -2 + \ln t \) represent a curve in the plane where both x and y change as \( t \) changes. This different perspective can often simplify complex problems, particularly when dealing with curves that are not easily described by a function \( y = f(x) \).

• Advantages: Parametric equations can represent more complex motion and trajectories, which are difficult to express using a single equation.
• Example: Imagine plotting the position of a bird that flies in a loop. Using parametric equations, we can represent its east-west and north-south positions separately by using time as a parameter.

Comprehending the Second Derivative

When we calculate the derivative, we are essentially looking at the rate of change or the slope of a curve at a particular point. The second derivative, denoted as \( \frac{d^2y}{dx^2} \), gives us more in-depth information—it tells us about the curvature, concavity, or how the slope itself is changing over time. In simpler terms, while the first derivative tells us about the direction and steepness of a curve, the second derivative reveals if the curve is curving upwards or downwards at any given point. If \( \frac{d^2y}{dx^2} > 0 \), the curve is concave up (like a cup), and if \( \frac{d^2y}{dx^2} < 0 \), it's concave down (like a dome).

• In the exercise: The second derivative \( \frac{d^2y}{dx^2} = t^2 \) was found. By substituting \( t = 1 \), \( \frac{d^2y}{dx^2} = 1 \), indicating that the curve is concave upwards at the point.
• Applications: The second derivative can be crucial in optimizing functions and determining maxima or minima, as well as for understanding the behavior of accelerating objects.

Decoding Differentiation in Parametric Form

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a quantity changes. In a parametric context, we differentiate each function separately with respect to \( t \) and then find the derivative \( \frac{dy}{dx} \) using the relationship between \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \). The calculation of \( \frac{dy}{dx} \) is done using the formula:\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]This ratio gives us the slope of the curve or the tangent line at any point \( t \). It essentially relates how quickly y changes compared to x as \( t \) varies.

• Example in the exercise: Given derivatives \( \frac{dx}{dt} = -\frac{1}{t^2} \) and \( \frac{dy}{dt} = \frac{1}{t} \), the expression \( \frac{dy}{dx} = -t \) was achieved, indicating the slope of the tangent at \( t=1 \) is \(-1\).
• Tip: Always remember to substitute the correct value of \( t \) to find the slope at the specific point of interest.