Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

, find the length of the parametric curve defined over the given interval. $$ x=\sqrt{1-t^{2}}, y=1-t ; 0 \leq t \leq \frac{1}{4} $$

Step-by-Step Solution

Answer

The length of the curve is approximately 0.2527.

Step 1: Understand the formula for arc length of a parametric curve

The formula for the length of a parametric curve defined by \( x(t) \) and \( y(t) \) from \( t=a \) to \( t=b \) is given by \[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]. We'll use this formula to find the length of the curve.

Step 2: Compute the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)

Differentiate \( x(t) = \sqrt{1-t^2} \) with respect to \( t \), yielding \( \frac{dx}{dt} = \frac{-t}{\sqrt{1-t^2}} \). Differentiate \( y(t) = 1-t \) with respect to \( t \), yielding \( \frac{dy}{dt} = -1 \).

Step 3: Substitute the derivatives into the arc length formula

Substitute \( \frac{dx}{dt} = \frac{-t}{\sqrt{1-t^2}} \) and \( \frac{dy}{dt} = -1 \) into the arc length formula: \[ L = \int_{0}^{1/4} \sqrt{\left( \frac{-t}{\sqrt{1-t^2}} \right)^2 + (-1)^2} \, dt \]. This simplifies to \[ L = \int_{0}^{1/4} \sqrt{\frac{t^2}{1-t^2} + 1} \, dt \].

Step 4: Simplify the integrand

Simplify the expression under the square root: \( \frac{t^2}{1-t^2} + 1 = \frac{t^2 + (1-t^2)}{1-t^2} = \frac{1}{1-t^2} \), so the integrand becomes \( \sqrt{\frac{1}{1-t^2}} = \frac{1}{\sqrt{1-t^2}} \).

Step 5: Evaluate the integral

Now, evaluate the integral: \( L = \int_{0}^{1/4} \frac{1}{\sqrt{1-t^2}} \, dt \). This is a standard integral form which is \( \sin^{-1}(t) + C \). Evaluate it from 0 to 1/4 to get \( \sin^{-1}(1/4) - \sin^{-1}(0) = \sin^{-1}(1/4) \).

Step 6: Find numerical approximation for \( \sin^{-1}(1/4) \)

Using a calculator, compute \( \sin^{-1}(1/4) \) which is approximately 0.2527.

Key Concepts

Understanding Arc Length

Arc length is a measure of the distance along a curve. In calculus, when you have a curve represented by a parametric function, you can calculate its length using a specific integral. This is particularly important because curves aren't always straight, and regular length measurements don't apply nicely.

The formula to calculate the arc length of a parametric curve is:

• If the curve is described by functions \(x(t)\) and \(y(t)\), then the arc length \(L\) from \(t = a\) to \(t = b\) can be found with:

\[L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt.\]This integral gives the exact distance travelled along the curve between two points, calculated by considering the rate of change of \(x\) and \(y\) with respect to \(t\).

By applying this formula, you can determine not just the simple straight-line distance between points, but the true distance along a potentially winding or intricate path.

Working with Parametric Curves

Parametric curves differ from traditional functions because they use an independent parameter—often \(t\)—to describe the coordinates \(x\) and \(y\) independently. This can model more complex curves that cannot be represented as a simple function \(y=f(x)\).

For instance, the parametric equations provided:

• \(x(t) = \sqrt{1-t^2}\)
• \(y(t) = 1-t\)

These define a curve segment where both \(x\) and \(y\) are related to the parameter \(t\), which ranges from 0 to \(\frac{1}{4}\). This gives each point on the curve a specific \(t\)-value based on the equations, generating both \(x\) and \(y\) coordinates.

Using parametric equations is helpful for situations where the curve loops, twists, or doubles back on itself, which can happen with physical trajectories, or certain geometrical shapes, like ellipses or circles.

Integral Evaluation

To compute the arc length or other quantities using calculus, many problems can break down to an integral evaluation. This process involves finding the integral of the function, which may sometimes require simplification or transformation for a proper solution.

In the case of finding the arc length for parametric curves, the integral setup involves the root function which accounts for changes in both \(x(t)\) and \(y(t)\). In our problem:

• After differentiating \(x(t)=\sqrt{1-t^2}\) and \(y(t)=1-t\), you derive expressions for \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
• Substituting these into the arc length formula gives a new integrand.
• Further simplification, as shown, reduces the complexity so the integral becomes easier to compute.

For many common forms, like \(\int \frac{1}{\sqrt{1-t^2}} \, dt\), you can use known results or transformations, such as inverse trigonometric functions. Here, the integral evaluates to \(\sin^{-1}(t)\), highlighting the connection between integration, geometry, and trigonometric identities. Evaluating from the parameter limits allows for solving the problem numerically or analytically to get the arc length.