Problem 99
Question
Distance A line with slope \(m\) passes through the point \((0,4) .\) (a) Write the shortest distance \(d\) between the line and the point \((3,1)\) as a function of \(m .\) (b) Use a graphing utility to graph the equation in part (a). (c) Find \(\lim _{m \rightarrow \infty} d(m)\) and \(\lim _{m \rightarrow-\infty} d(m) .\) Interpret the results geometrically.
Step-by-Step Solution
Verified Answer
The function of the shortest distance \(d(m)\) from the line \(y = mx + 4\) to the point \((3,1)\) is \(d(m) = | -3m - 3 | /\sqrt{m^2+1}\). The limit of this function as \(m\) approaches \(\pm \infty\) is 0. Geometrically, this represents that as the slope of the line becomes infinitely large (or negative), the line becomes more vertical, and the shortest distance to the point \((3,1)\) approaches 0.
1Step 1: Equation of the Line
Firstly, the equation of the line with slope \(m\) passing through the point \((0,4)\) is derived from the point-slope form of a linear equation \(y - y_1 = m(x - x_1)\). The given points substitute \(x_1\) and \(y_1\), giving \(y - 4 = m(x - 0)\) which simplifies to \(y = mx + 4\).
2Step 2: Distance from Point to Line Formula
The distance \(d\) from the point \((x_1, y_1) = (3,1)\) to the line with equation \(Ax + By + C = 0\) is given by the formula | \(Ax_1 + By_1 + C\) |/ sqrt(\(A^2 + B^2\)). Comparing this to the derived equation \(y - mx - 4 = 0\), \(A = -m\), \(B = 1\), and \(C = -4\). Substitution of these values into the formula gives \(d = | -m*3 + 1 - 4 | /\sqrt{(-m)^{2} + 1^{2}} = | -3m - 3 | /\sqrt{m^2+1}\).
3Step 3: Graphing the Function \(d(m)\)
Graph the absolute value function \(d(m) = | -3m - 3 | /\sqrt{m^2+1}\) using a graphing utility. Observe the basic shape and properties of the graph.
4Step 4: Computing Limits
For the limit of \(d(m)\) as \(m\) approaches \(\pm \infty\), the denominator increases faster than the numerator. These limits are both 0.
5Step 5: Geometric Interpretation
As the slope approaches \(\pm \infty\), the line becomes more vertical, so the shortest distance to the point approaches 0.
Key Concepts
Linear EquationsDistance FormulaLimit of a FunctionGraphing Utility
Linear Equations
A linear equation represents a straight line in a two-dimensional space. The standard form is given by \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. The slope \(m\) reflects how steep the line is, depicting how much \(y\) changes for each unit increase in \(x\).
In our exercise, the line has a slope \(m\) and passes through the point \((0, 4)\). By substituting this point into the point-slope form \(y - y_1 = m(x - x_1)\), we derive the line equation \(y = mx + 4\).
This formula allows us to understand how the line behaves as the slope varies. The key takeaway is how slope impacts the direction and steepness of the line.
In our exercise, the line has a slope \(m\) and passes through the point \((0, 4)\). By substituting this point into the point-slope form \(y - y_1 = m(x - x_1)\), we derive the line equation \(y = mx + 4\).
This formula allows us to understand how the line behaves as the slope varies. The key takeaway is how slope impacts the direction and steepness of the line.
Distance Formula
Calculating the distance from a point to a line involves a special formula. For a line in the form \(Ax + By + C = 0\), the distance \(d\) from a point \((x_1, y_1)\) is:
In the exercise, after deriving the line equation \(y = mx + 4\), we express it in standard form \(y - mx - 4 = 0\). Here, \(A = -m\), \(B = 1\), and \(C = -4\), enabling us to plug them into the distance formula.
The calculated distance function, \(d(m) = \frac{| -3m - 3 |}{\sqrt{m^2 + 1}}\), shows us how the distance between the line and the point \((3, 1)\) evolves as the slope \(m\) changes.
- \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
In the exercise, after deriving the line equation \(y = mx + 4\), we express it in standard form \(y - mx - 4 = 0\). Here, \(A = -m\), \(B = 1\), and \(C = -4\), enabling us to plug them into the distance formula.
The calculated distance function, \(d(m) = \frac{| -3m - 3 |}{\sqrt{m^2 + 1}}\), shows us how the distance between the line and the point \((3, 1)\) evolves as the slope \(m\) changes.
Limit of a Function
Limits help understand the behavior of a function as it approaches a specific point or infinity. In calculus, determining limits like \(\lim_{m \to \infty} d(m)\) reveals the end behavior.
In our case, the distance function \(d(m) = \frac{| -3m - 3 |}{\sqrt{m^2+1}}\) approaches a limit as \(m\) approaches infinity (\(m \to \pm \infty\)).
As solved previously, both limits at \(+\infty\) and \(-\infty\) are 0. This makes sense geometrically because as the slope \(m\) becomes extremely large or small, the line aligns more vertically, reducing the shortest distance to the point to zero.
Limits provide insight into how functions behave asymptotically or at boundaries.
In our case, the distance function \(d(m) = \frac{| -3m - 3 |}{\sqrt{m^2+1}}\) approaches a limit as \(m\) approaches infinity (\(m \to \pm \infty\)).
As solved previously, both limits at \(+\infty\) and \(-\infty\) are 0. This makes sense geometrically because as the slope \(m\) becomes extremely large or small, the line aligns more vertically, reducing the shortest distance to the point to zero.
Limits provide insight into how functions behave asymptotically or at boundaries.
Graphing Utility
Graphing utilities are tools that allow us to visualize functions and their behaviors. By plotting the function \(d(m) = \frac{| -3m - 3 |}{\sqrt{m^2+1}}\), we can directly observe patterns rather than just calculating numbers.
When we graph this function, we see the shape and symmetry, including its behavior as \(m\) tends to \(\pm \infty\). Visualizing the graph clarifies how the distance fluctuates with different slopes \(m\).
Graphing is essential in calculus and linear algebra to understand complex interactions and verify results. It enhances comprehension by transforming abstract numbers into tangible curves and shapes.
When we graph this function, we see the shape and symmetry, including its behavior as \(m\) tends to \(\pm \infty\). Visualizing the graph clarifies how the distance fluctuates with different slopes \(m\).
Graphing is essential in calculus and linear algebra to understand complex interactions and verify results. It enhances comprehension by transforming abstract numbers into tangible curves and shapes.
Other exercises in this chapter
Problem 97
Proof In Exercises \(95-98\) , use the definition of limits at infinity to prove the limit. $$ \lim _{x \rightarrow-\infty} \frac{1}{x^{3}}=0 $$
View solution Problem 98
Proof In Exercises \(95-98\) , use the definition of limits at infinity to prove the limit. $$ \lim _{x \rightarrow-\infty} \frac{1}{x-2}=0 $$
View solution Problem 99
Proof Use the definitions of increasing and decreasing functions to prove that \(f(x)=x^{3}\) is increasing on \((-\infty, \infty) .\)
View solution Problem 100
Distance \(A\) line with slope \(m\) passes through the point \((0,-2) .\) (a) Write the shortest distance \(d\) between the line and the point \((4,2)\) as a f
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