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Magazine Chapter 4: Problem 16
Find the point on the curve \(y=\sqrt{x}\) that is closest to the point \((3,0) .\)
Short Answer
Expert verified
The closest point is \(\left(\frac{5}{2}, \sqrt{2.5}\right)\).
Step by step solution
01
Understand the Problem
We need to find the point on the curve \(y = \sqrt{x}\) that is closest to the point \((3, 0)\). The closest distance can be found using the distance formula and minimizing that expression.
02
Express Distance Formula
The distance \(d\) between a point \((x, \sqrt{x})\) on the curve and the point \((3, 0)\) is given by the distance formula: \[d = \sqrt{(x - 3)^2 + (\sqrt{x} - 0)^2} = \sqrt{(x - 3)^2 + x}.\]
03
Simplify the Problem
To make it easier for optimization, we can minimize the square of the distance (since the square root is a monotonic function, the minimum distance will correspond to the minimum of the squared distance).\[f(x) = (x - 3)^2 + x.\]
04
Differentiate the Function
Find the derivative of \(f(x)\) to locate its critical points. \[f'(x) = 2(x - 3) + 1 = 2x - 6 + 1 = 2x - 5.\]
05
Solve for Critical Points
Set \(f'(x) = 0\) to find critical points. \[2x - 5 = 0\]Solving for \(x\), we get \[x = \frac{5}{2}.\]
06
Determine the Closest Point
Substitute \(x = \frac{5}{2}\) into the original equation of the curve to find the corresponding \(y\)-value. \[y = \sqrt{\frac{5}{2}} = \sqrt{2.5}.\]Thus, the point is \(\left(\frac{5}{2}, \sqrt{2.5}\right)\).
07
Verify the Minimum
Verify whether the point \((\frac{5}{2}, \sqrt{2.5})\) is indeed a minimum by checking the second derivative of \(f(x)\). \[f''(x) = 2.\]Since \(f''(x) > 0\), the function \(f(x)\) is concave up and thus has a local minimum at \(x = \frac{5}{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Distance Formula
The distance formula helps us find the distance between two points in a coordinate plane. In this exercise, our goal is to measure how close a point on the curve \(y = \sqrt{x}\) is to the fixed point \((3,0)\). The distance formula is given by:
- \(\sqrt{(x - x_1)^2 + (y - y_1)^2}\)
- \(d = \sqrt{(x - 3)^2 + (\sqrt{x} - 0)^2}\)
Derivative
Derivatives are powerful tools in calculus that represent how a function changes at any point, showing the rate at which something happens. In optimization, we take the derivative to find critical points.
In our function \(f(x) = (x - 3)^2 + x\), we find the derivative \(f'(x)\), which gives us:
In our function \(f(x) = (x - 3)^2 + x\), we find the derivative \(f'(x)\), which gives us:
- \(f'(x) = 2(x - 3) + 1\)
- This simplifies to \(f'(x) = 2x - 5\)
Critical Points
A critical point occurs where a function's derivative equals zero or does not exist. These points are potential locations for local maxima, minima, or saddle points.
For our function \(f(x) = (x - 3)^2 + x\), setting the first derivative equal to zero, \(f'(x) = 2x - 5 = 0\), helps us find the critical point:
For our function \(f(x) = (x - 3)^2 + x\), setting the first derivative equal to zero, \(f'(x) = 2x - 5 = 0\), helps us find the critical point:
- \(x = \frac{5}{2}\)
Second Derivative Test
The second derivative test helps us determine the nature of a critical point by examining the sign of the second derivative at that point.
Simply put, if the second derivative is positive at a critical point, the function is concave up, indicating a local minimum. Conversely, if negative, it's concave down, indicating a local maximum.
Simply put, if the second derivative is positive at a critical point, the function is concave up, indicating a local minimum. Conversely, if negative, it's concave down, indicating a local maximum.
- For \(f(x) = (x - 3)^2 + x\), the second derivative: \(f''(x) = 2\)
- Since \(f''(x) > 0\), we conclude the critical point \(x = \frac{5}{2}\) is indeed a minimum.
