Problem 28 - Calculus Early Transcendentals

Question

What curve passes through the point (1,5) and has an arc length on the interval [2,6] given by \(\int_{2}^{6} \sqrt{1+16 x^{-6}} d x ?\)

Step-by-Step Solution

Answer

Answer: The equation of the curve is \(y(x) = -2x^{-2} + 7\).

Step 1: Identify the arc length formula

The arc length formula for a curve y(x) is given by: \(L = \int_{a}^{b} \sqrt{1 + (y'(x))^2} dx\) In our case, the given integral represents the arc length, and we have: \(\int_{2}^{6} \sqrt{1+16 x^{-6}} dx = L\)

Step 2: Find the derivative of the curve

According to the arc length formula, we have \(\sqrt{1 + (y'(x))^2} = \sqrt{1 + 16x^{-6}}\). Let's find the derivative of the curve, y'(x), by squaring both sides and simplifying: \((1 + (y'(x))^2) = 1 + 16x^{-6}\) \((y'(x))^2 = 16x^{-6}\) \(y'(x) = 4x^{-3}\) (we only consider the positive root because the curve has a positive arc length)

Step 3: Integrate the derivative to find the curve equation

Now that we have the derivative, integrate it to find the curve equation y(x): \(y(x) = \int y'(x) dx = \int 4x^{-3} dx = -2x^{-2} + C\), where C is the integration constant

Step 4: Use the given point to solve for the integration constant

Given the curve passes through the point (1,5), plug in the coordinates into the curve equation to find C: \(5 = -2(1)^{-2} + C\) \(5 = -2 + C\) \(C = 7\)

Step 5: Write the final curve equation

Now, substitute the value of C back into the curve equation to get the final equation for the curve: \(y(x) = -2x^{-2} + 7\) Thus, the curve passing through the point (1,5) and having an arc length on the interval [2,6] given by the integral function is \(y(x) = -2x^{-2} + 7\).

Key Concepts

Integral Calculus

Integral Calculus is a branch of mathematics focused on integrations. It's crucial for calculating areas under curves, volumes, and in this case, arc lengths of curves.
The core idea is to accumulate small quantities over an interval to find a total. In this exercise, the arc length is given by the integral: \[\int_{2}^{6} \sqrt{1+16 x^{-6}} \, dx\] Here, the integrand, \(\sqrt{1+16 x^{-6}}\), represents the element of arc length, which needs to be summed up from \(x=2\) to \(x=6\).

• Integral defines the function of accumulated lengths over an interval by evaluating the area that the curve underlies in its path.
• Integral calculus thus connects derivatives and integrals, serving as inverse operations to each other.

Makes sure to practice integration as it's vital for finding exact descriptions of curves, like in the arc length formula here.

Curve Equation

The curve equation describes the curve's path and behavior in a coordinate system. This involves finding a function \(y(x)\) that satisfies all given conditions.
In our problem, the goal is to find a curve that an integral's arc length represents.The curve's derivative \(y'(x)\) is initially derived from the arc length formula: \[y'(x) = 4x^{-3}\] Next, you integrate this derivative to get the curve equation: \[y(x) = \int 4x^{-3} \, dx = -2x^{-2} + C\] This is the general form of the curve solution, depending on the constant \(C\) to provide specific solutions.

• The curve equation can guide you in predicting how a curve will change over different domains.
• Identifying the correct curve equation can solve various problems involving paths, motion, and rates of change.

The curve in this problem scenario is particular to the arc length from the integral and the exact path point \((1,5)\).

Differential Calculus

Differential Calculus deals with the concept of derivatives, which represents rates of change of functions.
In the context of the problem, it helps us understand the curve's gradient, slope or rate of change at any point.The arc length expression gave us: \[\sqrt{1 + (y'(x))^2} = \sqrt{1 + 16x^{-6}}\] By setting these equal, and solving for the derivative, it guides us to find: \[y'(x) = 4x^{-3}\]

• This derivative notion is a representation of the original curve's slope at any point \(x\).
• The arc length component squared becomes a simple expression representing the derivative squared.

Differential calculus shines in predicting how functions behave and transform, and in this case, helps finalize the proper derivative needed for integration.

Integration Constant

The integration constant \(C\) is a vital part of indefinite integrals. It represents the unknown constant value which ensures the integrated function can fit any initial or boundary condition.
Our integral of the derivative was: \[y(x) = -2x^{-2} + C\] The constant \(C\) gets determined using a specific known point of the curve, \((1,5)\). Plug the coordinates into the equation to solve for \(C\):

• \(5 = -2(1)^{-2} + C\)
• \(5 = -2 + C\)
• Therefore, \(C = 7\)

Reincorporating this constant into the curve equation provides the complete description \(y(x) = -2x^{-2} + 7\). Integrals often account for numerous possible solutions, but specific conditions like this are what pinpoint a single, exact curve solution.