Question

$$ \begin{array}{c}{\text { (a) Group Activity Find a curve through the point }(1,1)} \\ {\text { whose length integral is } y=\sqrt{x}} \\\ {L=\int_{1}^{4} \sqrt{1+\frac{1}{4 x}} d x}\end{array} $$ $$ \begin{array}{l}{\text { (b) Writing to Learn How many such curves are there? Give }} \\ {\text { reasons for your answer. }}\end{array} $$

Short answer

The curve passing through the point (1,1) and satisfying the given length integral is \(y=\ln(2x)\). Only one such curve exists.

Step-by-step solution

Set up Integral

First, set up the length integral \(L=\int_{1}^{4} \sqrt{1+\frac{1}{4 x}} dx\).

Solve Integral

This integral is a standard one, which can be resolved using a simple u-substitution method. By setting \(u=2\sqrt{x}\), the equation \(L\) becomes: \(L=\int_{\sqrt{2}}^{2\sqrt{2}} \frac{du}{u}\). Solving this integral gives \(L=\ln(u)\Big|_{\sqrt{2}}^{2\sqrt{2}}\). Evaluating at these limits yields \(L=\ln(2\sqrt{2})-\ln(\sqrt{2})=\ln(2)\).

Define Curve Function

With \(L=\ln(2)\), equate \(L\) to the curve length, \(y=\sqrt{x}\) at point (1,1). Therefore, the required curve must satisfy \(y=\sqrt{x}=\ln(2)\). Since we had a specific point (1,1), the actual curve is \(y=\ln(2x)\).

Reasoning About Curve Selection

Regarding the number of possible curves, there's only one curve satisfying the length integral. Any other curve passing through the point (1,1) will not validate the length integral. Moreover, arithmetic integral operations ensure unique solutions, so there's only one such curve.

Key concepts

U-Substitution Method

The u-substitution method is a powerful tool in integral calculus, particularly helpful when dealing with problematic integrals. It involves changing the variable of integration to simplify the integral and make it more manageable. This method is often compared to the reverse process of the chain rule in differentiation.

To implement u-substitution, we choose a substitution variable, often denoted by 'u', which is a function of the original variable, say 'x'. The differential 'du' is then expressed in terms of 'dx', which allows us to rewrite the integral in terms of 'u'. After integrating with respect to 'u', we substitute back the original variable 'x' to find the solution.

For example, in our exercise, by setting \( u = 2\sqrt{x} \), and then expressing 'du' in terms of 'dx', the integral \( L=\int_{1}^{4} \sqrt{1+\frac{1}{4x}} dx \) becomes simpler and expressed as \( L=\int_{\sqrt{2}}^{2\sqrt{2}} \frac{du}{u} \), which is a straightforward integral to evaluate.

Definite Integrals

Definite integrals are foundational in calculus, representing the net area under a curve on a specific interval. This interval is defined by two boundaries, often labeled as the lower limit 'a' and the upper limit 'b'. The definite integral accumulates the difference between the values of the integral at these boundaries.

Symbolically, we denote a definite integral as \( \int_{a}^{b} f(x) dx \), where \( f(x) \) is the function we're integrating over the interval \( [a, b] \). When resolving a definite integral, we find the antiderivative of \( f(x) \), usually denoted by 'F(x)', and then apply the evaluation theorem, \( F(b) - F(a) \), which gives us the net area under the curve between 'a' and 'b'.

In the provided exercise, after employing the u-substitution method, we get to the step \( L=\ln(u)\Big|_{\sqrt{2}}^{2\sqrt{2}} \). Here, we're dealing with a definite integral where the antiderivative of \( \frac{1}{u} \) is \( \ln(|u|) \), and the evaluation theorem is used to find the definite answer.

Calculus Curve Lengths

Calculating the length of a curve in calculus involves a specific application of integral calculus, often referred to as the arc length. To find the length of a curve represented by a function 'y = f(x)' over an interval 'a' to 'b', we use the length integral formula:
\[ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \].

This formula derives from the Pythagorean theorem applied to an infinitesimally small segment of a curve. By summing up these segments over the intended interval, we obtain the total length of the curve.

In our exercise, the length integral \( L=\int_{1}^{4} \sqrt{1+\frac{1}{4x}} dx \) corresponds to the length of the specific curve through the point (1,1). The integral is solved using the u-substitution method, giving us a definite integral. We find that the curve with the length integral as \( y=\sqrt{x} \) is unique; any other curve with the same length won't necessarily pass through the point (1,1). Hence, there is only one curve that satisfies both the curve's length and the requirement to pass through the given point.