Question

Proximity questions Find the point on the graph of \(y=\sqrt{x}\) that is nearest the point \((p, 0)\) if (i) \(p>\frac{1}{2} ;\) and (ii) \(0

Short answer

Question: Find the point on the graph of \(y = \sqrt{x}\) that is nearest to a point \((p, 0)\) on the x-axis, considering two cases: (i) when \(p > \frac{1}{2}\), and (ii) when \(0 < p < \frac{1}{2}\). Answer: (i) For the case \(p > \frac{1}{2}\), the nearest point to \((p, 0)\) on the graph of \(y = \sqrt{x}\) is at \(x = p - \frac{1}{2}\) and \(y = \sqrt{p-\frac{1}{2}}\). (ii) For the case \(0 < p < \frac{1}{2}\), the nearest point to \((p, 0)\) on the graph of \(y = \sqrt{x}\) is at \(x = 0\) and \(y = 0\).

Step-by-step solution

Write down the distance formula

Recall the distance formula for two points \((x_1, y_1)\) and \((x_2, y_2)\) in the plane: $$ d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} $$ We will use this formula to find the distance between any point \((x, y)\) on the graph of \(y = \sqrt{x}\) and point \((p, 0)\).

Apply the distance formula to the given points

Let \((x, y)\) be a point on the graph of \(y = \sqrt{x}\). Then the distance between \((x, y)\) and \((p, 0)\) is given by: $$ d(x) = \sqrt{(x-p)^2+(\sqrt{x}-0)^2} $$ Simplify the expression: $$ d(x) = \sqrt{(x-p)^2+x} $$ Our task is to minimize this distance function \(d(x)\) for both cases.

Differentiate the distance function with respect to x

To find the minimum value of the distance function, we will differentiate it with respect to \(x\) and then set the derivative equal to zero to find the critical points. We will need to use the chain rule for differentiation: $$ \frac{d}{dx}[d(x)] = \frac{d}{dx}\left[\sqrt{(x-p)^2+x}\right] $$ Let \(u=(x-p)^2+x\). Then \(d(x)=\sqrt{u}\) and to differentiate \(d(x)\) with respect to x, we need to differentiate \(u\) with respect to x: $$ \frac{du}{dx} = \frac{d}{dx}\left[(x-p)^2+x\right] = 2(x-p)+1 $$ Now, apply the chain rule: $$ \frac{d}{dx}[d(x)] = \frac{1}{2\sqrt{u}}\left(\frac{du}{dx}\right) = \frac{1}{2\sqrt{(x-p)^2+x}}\left(2(x-p)+1\right) $$

Set the derivative equal to zero and solve for x

To find the critical points, set the derivative equal to zero and solve for x: $$ \frac{1}{2\sqrt{(x-p)^2+x}}\left(2(x-p)+1\right) = 0 $$ The fraction is equal to zero when its numerator is equal to zero, so we have: $$ 2(x-p)+1 = 0 $$ Solve for \(x\): $$ x = p - \frac{1}{2} $$

Analyze the critical point for each case

Now, let's analyze the critical point for each case: (i) \(p > \frac{1}{2}\): In this case, \(0 < p - \frac{1}{2}\). Therefore, the point \((x, y)\) on the graph of \(y = \sqrt{x}\) that is nearest to \((p, 0)\) is given by: $$ x = p - \frac{1}{2} \quad \text{and} \quad y = \sqrt{x} = \sqrt{p-\frac{1}{2}} $$ (ii) \(0 < p < \frac{1}{2}\): In this case, \(p - \frac{1}{2}<0\). But \(x\) cannot be negative since \(y = \sqrt{x}\), so for this case, the point on the graph nearest to \((p, 0)\) is at \(x=0\). Therefore, the point \((x, y)\) is given by: $$ x = 0 \quad \text{and} \quad y = \sqrt{x} = 0 $$

Key concepts

Distance Formula

The distance formula is a crucial tool in geometry and calculus for determining the distance between two points in a coordinate plane. Consider two points \((x_1, y_1)\) and \((x_2, y_2)\). The distance \(d\) between them is calculated with the formula: \\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\]Here's how this works:

• Calculate the horizontal and vertical differences between the points: \(x_2 - x_1\) and \(y_2 - y_1\).
• Square each difference to eliminate negative values.
• Add these squares together.
• Take the square root to find the distance.

In our exercise, this formula helps find the point on the curve \(y = \sqrt{x}\) closest to a given point \((p, 0)\). Understanding how to apply the distance formula is essential for solving proximity problems effectively.

Graph of a Function

Graphs visually represent relationships between variables. For the function \(y = \sqrt{x}\), the graph is a curve starting at the origin and rising gradually to the right.
Understanding this graph is key to solving proximity problems as it helps in identifying where points lie.
Key Characteristics of \(y = \sqrt{x}\) Graph:

• Domain: \(x \geq 0\). The square root function is not defined for negative numbers, which affects proximity calculations when \(x\) can't be negative.
• Range: \(y \geq 0\). The function only gives non-negative \(y\) values.
• Shape: The graph rises slower at larger \(x\) since the rate of change decreases.

By understanding the graph's domain and range, you can determine feasible solutions for points closest to other specified coordinates.

Critical Points

Critical points are values of \(x\) where the derivative of a function is zero or undefined. They help find maximum or minimum points on a graph.
In proximity problems, finding critical points can help determine the nearest point on a graph to a given point.
Finding Critical Points:

• Find the derivative of the function.
• Set the derivative equal to zero.
• Solve for \(x\) to find potential critical points.
• Analyze these points to identify minimum or maximum values.

For the function \(d(x) = \sqrt{(x-p)^2+x}\), calculating the derivative and setting it to zero finds the critical points. These points indicate where the distance is minimized relative to \((p, 0)\). This approach helps understand how the graph behaves around different points on the \(x\)-axis.

Chain Rule

The chain rule is a fundamental technique in calculus, essential for finding derivatives of composite functions.
It simplifies finding the derivative when one function is nested inside another.
Using the Chain Rule:

• Identify the inner (\(u\)) and outer functions.
• Differentiate both functions separately.
• The derivative of the composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.
• Formula: \\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]

In our proximity problem, to differentiate \(d(x) = \sqrt{(x-p)^2+x}\), identify \(u = (x-p)^2+x\) and \(d(x) = \sqrt{u}\). The chain rule helps differentiate these efficiently, leading to the solution for closest points.