Problem 1 - Thomas Calculus

Question

Find the lengths of the curves in Exercises \(1-6\) $$ x=1-t, \quad y=2+3 t, \quad-2 / 3 \leq t \leq 1 $$

Step-by-Step Solution

Answer

The length of the curve is \( \frac{5 \sqrt{10}}{3} \).

Step 1: Parametric Function Setup

We have the parametric equations for the curve given by \( x = 1 - t \) and \( y = 2 + 3t \). The task is to find the length of the curve over the interval \( -\frac{2}{3} \leq t \leq 1 \).

Step 2: Find Derivatives

To find the length of the curve, we first need to compute the derivatives of the parametric equations. The derivative \( \frac{dx}{dt} = -1 \) and \( \frac{dy}{dt} = 3 \).

Step 3: Use Curve Length Formula

The formula for the length of a curve defined by parametric equations \( x = f(t) \) and \( y = g(t) \) over interval \([a, b]\) is given by: \[ L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]Substitute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) into this formula.

Step 4: Integral Setup

Substitute the derivatives into the length formula to get: \[ L = \int_{-2/3}^{1} \sqrt{(-1)^2 + (3)^2} \, dt \] This simplifies to: \[ L = \int_{-2/3}^{1} \sqrt{1 + 9} \, dt \] \[ L = \int_{-2/3}^{1} \sqrt{10} \, dt \].

Step 5: Evaluate the Integral

The integral \(L = \int_{-2/3}^{1} \sqrt{10} \, dt \) becomes \[ L = \sqrt{10} \int_{-2/3}^{1} \, dt \].Evaluate the definite integral: \[ \int_{-2/3}^{1} \, dt = [t]_{-2/3}^{1} = 1 - (-2/3) = 1 + 2/3 = \frac{5}{3} \].

Step 6: Calculate the Curve Length

Now multiply the result of the integral evaluation by \( \sqrt{10} \): \[ L = \sqrt{10} \cdot \frac{5}{3} = \frac{5 \sqrt{10}}{3} \].

Key Concepts

Parametric Equations

Parametric equations allow you to express a curve using a parameter, typically represented as "t". These equations define the x and y coordinates uniquely in terms of "t". In this exercise, the parametric equations are given by

• \( x = 1 - t \)
• \( y = 2 + 3t \)

These two equations describe a straight line in a plane. Rather than expressing \( y \) directly in terms of \( x \), parametric equations provide a versatile way to portray complex curves beyond simple functions.
In this specific problem, the parameter "t" ranges from \(-\frac{2}{3} \) to \(1\), which identifies which portion of the line you're interested in studying.
Understanding these equations can unlock insights into the geometry and behavior of curves and allow for more flexible problem-solving techniques than traditional Cartesian functions.

Integral Calculus

Integral calculus plays a vital role when it comes to measuring properties such as area, volume, and in this case, the length of curves. The length of a curve defined via parametric equations can be determined using the integral curve length formula:\[L = \int_a^b \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]
This formula essentially adds up infinitely small line segments dictated by the changes in \( x \) and \( y \) as dictated by their derivatives, forming the entire curve over the interval \([a, b]\).
In the given solution, the integral of \( \int_{-2/3}^{1} \sqrt{10} \ dt \) calculates the total length by considering all the small changes over "t" from \(-\frac{2}{3} \) to \(1\). Calculus integrates these components to provide a collective measure, demonstrating the power of integration in solving real-world problems such as finding the arc length.

Derivatives

Derivatives represent the rate at which things change. For parametric equations, we compute the derivative of \( x \) and \( y \) with respect to the parameter "t", which provides details on how each coordinate changes as "t" changes.
In the provided exercise, the derivatives are:

• \( \frac{dx}{dt} = -1 \)
• \( \frac{dy}{dt} = 3 \)

These derivatives tell us the direction and speed of change for the curve in the x and y directions. For the length of the curve, they are squared, summed, and square-rooted in the curve length formula. Derivatives transform raw positions into information about how a point moves along the curve, helping us piece together the whole shape and path of the curve.