Chapter 1: Problem 14
Consider the point \((x, y)\) lying on the graph of \(y=\sqrt{x-3}\) . Let \(L\) be the distance between the points \((x, y)\) and \((4,0) .\) Write \(L\) as a function of \(y .\)
Short Answer
Expert verified
The function is \( L(y) = \sqrt{1 - y^2 + y^4} \).
Step by step solution
01
Express y in terms of x
The function given is \( y = \sqrt{x-3} \). We need to express \( x \) in terms of \( y \) to substitute it in the distance formula. Squaring both sides of the equation gives us \( x = y^2 + 3 \).
02
Write the distance formula
The distance \( L \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula: \( L = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). In our case, the points are \((x, y)\) and \((4, 0)\).
03
Substitute the points in the distance formula
Using the points \((x, y)\) and \((4, 0)\), the distance \( L \) is: \( L = \sqrt{(4 - x)^2 + (0 - y)^2} = \sqrt{(4 - x)^2 + y^2} \).
04
Replace x with expression in y
From Step 1, we have \( x = y^2 + 3 \). Substitute this into the distance formula: \( L = \sqrt{(4 - (y^2 + 3))^2 + y^2} \).
05
Simplify the expression
Simplify \( (4 - (y^2 + 3)) \) to \( 1 - y^2 \). Therefore, \( L = \sqrt{(1 - y^2)^2 + y^2} \). Expanding the square gives \( L = \sqrt{1 - 2y^2 + y^4 + y^2} = \sqrt{1 - y^2 + y^4} \).
06
Final Expression for L as a function of y
The distance \( L \, \) in terms of \( y \), is given by \( L(y) = \sqrt{1 - y^2 + y^4} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Function of a Variable
Understanding a function of a variable is vital when dealing with mathematical expressions. In our exercise, the relation for the point \( (x, y) \) is defined by the equation \( y = \sqrt{x-3} \). This indicates that \( y \) is a function of \( x \). In order to switch \( x \) to a function of \( y \), we employed algebraic manipulation. By squaring both sides of the equation, we solved for \( x \) and expressed it as \( x = y^2 + 3 \). Converting between functions like this is essential when adapting formulas to different points of reference, which is a common need in geometry and calculus. This manipulation allows us to then use this expression to find the distance between points as a function of \( y \).
Square Roots
Square roots are an essential mathematical concept used for solving equations and understanding relationships between variables. In the function \( y = \sqrt{x-3} \), we use the square root to define \( y \) in terms of \( x \). It helps in creating a curve on a graph representing all possible points \( (x, y) \) that satisfy the equation.When understanding square roots:
- It's the inverse operation of squaring a number.
- Square roots are always non-negative in real number contexts.
- They appear often when dealing with circles, parabolas, and distance problems.
Distance Between Points
Calculating the distance between two points on a graph is a fundamental skill in geometry and trigonometry. In this exercise, we use the distance formula to find the distance \( L \) between two points, \( (x, y) \) and \( (4, 0) \).The distance formula is given as:\[ L = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]In our exercise, this formula morphs when an expression for \( x \) in terms of \( y \) is substituted:
- First, we account for how \( x \) is expressed via \( y \): \( x = y^2 + 3 \).
- Substituting, we then derive \( L = \sqrt{1 - y^2 + y^4} \).