dy/dx=x/(x-4)
I.will have a slope field with negative slopes in quadrant I
II.will have a slope field with positive slopes in all quadrants
III.will produce a slope field with columns of parallel tangents
for x > 4, x/(x-4) > 1, so the slopes will be positive, leveling out
for x < 4, the slopes will be negative, leveling out.
The slopes will be approaching vertical at x=4.
Looks like III.
See
http://www.wolframalpha.com/input/?i=vector+plot&rawformassumption={%22F%22,+%22VectorPlot%22,+%22vectorfunction%22}+-%3E%22{x,x-4}%22&rawformassumption={%22F%22,+%22VectorPlot%22,+%22vectorplotvariable1%22}+-%3E%22x%22&rawformassumption={%22F%22,+%22VectorPlot%22,+%22vectorplotlowerrange1%22}+-%3E%22-10%22&rawformassumption={%22F%22,+%22VectorPlot%22,+%22vectorplotupperrange1%22}+-%3E%2210%22&rawformassumption={%22F%22,+%22VectorPlot%22,+%22vectorplotvariable2%22}+-%3E%22y%22&rawformassumption={%22F%22,+%22VectorPlot%22,+%22vectorplotlowerrange2%22}+-%3E%22-10%22&rawformassumption={%22F%22,+%22VectorPlot%22,+%22vectorplotupperrange2%22}+-%3E%2210%22&rawformassumption={%22MC%22,+%22%22}+-%3E+{%22Calculator%22}&rawformassumption={%22MC%22,%22%22}-%3E{%22Formula%22}
I. Oh no! The slopes in quadrant I will actually be positive. It seems like someone flipped the signs without asking for directions!
II. You got it! The slopes in the slope field will indeed be positive in all quadrants. I guess this function doesn't discriminate!
III. Oops! It's more like rows of parallel tangents rather than columns. Looks like the tangents are forming quite the queue, waiting for their turn to shine!
To determine the characteristics of the slope field of the differential equation dy/dx = x / (x - 4), we can analyze the equation and its behavior.
First, let's find the critical points by setting the denominator equal to zero:
x - 4 = 0
x = 4
So, the critical point is at x = 4.
Now, let's analyze the options given:
I. A slope field with negative slopes in quadrant I
To determine the direction of the slopes, consider substituting x values into the equation:
For x < 4, let's choose x = 0:
dy/dx = 0 / (0 - 4) = 0/(-4) = 0
For x > 4, let's choose x = 6:
dy/dx = 6 / (6 - 4) = 6/2 = 3
Therefore, for x < 4, dy/dx = 0, and for x > 4, dy/dx > 0. This means that the slopes are positive in quadrant I, NOT negative. So, option I is incorrect.
II. A slope field with positive slopes in all quadrants
As discussed above, dy/dx is positive for x > 4. Since the positive slopes occur when x > 4, they do not exist in all quadrants. Hence, option II is also incorrect.
III. A slope field with columns of parallel tangents
To determine if the slope field produces columns of parallel tangents, we need to consider the behavior of the slopes as x approaches 4 from both sides.
For x < 4, let's choose x = 3.9:
dy/dx = 3.9 / (3.9 - 4) = 3.9 / (-0.1) = -39
For x > 4, let's choose x = 4.1:
dy/dx = 4.1 / (4.1 - 4) = 4.1 / 0.1 = 41
As x gets closer to 4 from both sides, the slopes become extremely large or tend towards infinity. However, the slopes on opposite sides of x = 4 have the same sign (+41 and -39), which indicates that the tangents are not parallel. Thus, option III is also incorrect.
In summary, none of the given options accurately describe the characteristics of the slope field for the differential equation dy/dx = x / (x - 4).
To determine which statement(s) about the slope field of the differential equation dy/dx = x/(x-4) are true, we can analyze the equation and map out the slopes at various points.
First, let's find the critical points of the differential equation by setting the denominator equal to zero:
x - 4 = 0
Solving for x, we find that x = 4 is a critical point.
Now, let's consider each statement:
I. "The slope field will have negative slopes in quadrant I."
To determine this, we can evaluate the sign of the slope for points in quadrant I. If the derivative (slope) is negative at these points, then the statement is true.
For x > 4 (in quadrant I), the numerator (x) is positive, and the denominator (x-4) is positive. Therefore, the ratio x/(x-4) is positive. So, the derivative is positive in quadrant I. Hence, this statement is false.
II. "The slope field will have positive slopes in all quadrants."
To determine this, we can evaluate the sign of the slope for points in each quadrant. If the derivative (slope) is positive at all points, then the statement is true.
For x ≠ 4 (in all quadrants), the numerator (x) and denominator (x-4) have the same sign. Therefore, the ratio x/(x-4) is positive for all x ≠ 4, indicating that the derivative is positive. Thus, this statement is true.
III. "The slope field will produce columns of parallel tangents."
This statement suggests that there are vertical tangent lines (infinite slopes) for specific x-values. To determine if this is true, we need to check if there are any vertical asymptotes or singularities at certain x-values.
From the critical point analysis, we found that x = 4 is a critical point. By substituting x = 4 into the equation, we see that the denominator becomes zero. This implies that there is a singularity at x = 4.
When a singularity occurs, the slope field will not exhibit columns of parallel tangents. Instead, it may show a gap or break in the slope field at that particular x-value. Hence, this statement is false.
In summary:
- Statement I is false.
- Statement II is true.
- Statement III is false.