Sketching Slope Fields - Calculus 2

1. Consider the slope field for $\frac{dy}{dx} = \sin(x) + \sin(y)$. At which point in the square region $[0, 2\pi] \times [0, 2\pi]$ is the slope maximized?
   Answer: $(\pi/2, \pi/2)$ To maximize the slope $\frac{dy}{dx} = \sin(x) + \sin(y)$, we need to maximize both $\sin(x)$ and $\sin(y)$ independently. The maximum value of the sine function is 1. In the interval $[0, 2\pi]$, $\sin(x)=1$ occurs at $x=\pi/2$, and $\sin(y)=1$ occurs at $y=\pi/2$. Therefore, the maximum slope is $1+1=2$, and it occurs at the point $(\pi/2, \pi/2)$.
2. Consider the differential equation $\frac{dy}{dx} = (y^2 - 4)(x-1)$. A solution curve passes through the point $(1, 1)$. What is the nature of the solution curve at this point?
   Answer: The solution curve has a local maximum at $x=1$. First, evaluate the slope at $(1,1)$: $\frac{dy}{dx} = (1^2-4)(1-1) = (-3)(0) = 0$. This indicates a critical point. To classify it, we examine the sign of $\frac{dy}{dx}$ around $x=1$. For a point near $(1,1)$, $y$ is close to 1, so $y^2-4$ is negative. If $x > 1$, then $x-1 > 0$, so $\frac{dy}{dx} = (\text{neg})(\text{pos}) < 0$. If $x < 1$, then $x-1 < 0$, so $\frac{dy}{dx} = (\text{neg})(\text{neg}) > 0$. Since the derivative is positive to the left of $x=1$ and negative to the right, the solution curve has a local maximum at $x=1$.
3. Which of the following statements is true about the slope field for the differential equation $\frac{dy}{dx} = \sqrt{y - x^2}$?
   Answer: The slope field is defined only for points on or above the parabola $y=x^2$. The expression $\sqrt{y - x^2}$ is only defined for real numbers when the term under the square root is non-negative. Therefore, we must have $y - x^2 \ge 0$, which is equivalent to $y \ge x^2$. This inequality describes the set of points that are on or above the parabola $y=x^2$. The slope field is not defined for any points below this parabola. Choice B is incorrect because the field is undefined, not negative. Choice C is incorrect because slopes are zero on the parabola $y=x^2$, not the x-axis. Choice D is incorrect because the isoclines are other parabolas of the form $y=x^2+C^2$.
4. Consider the autonomous differential equation $\frac{dy}{dx} = \sin(\pi y)$. Which statement describes the line segments of the slope field in a small neighborhood of the line $y=1$?
   Answer: They point towards the line $y=1$, indicating a stable equilibrium. The line $y=1$ is an equilibrium solution because $\frac{dy}{dx} = \sin(\pi \cdot 1) = 0$. To determine its stability, we check the sign of $\frac{dy}{dx}$ nearby. For $y$ slightly greater than 1 (e.g., $y=1.1$), $\pi y$ is slightly greater than $\pi$, so $\sin(\pi y) < 0$. The solution curve moves downward, toward $y=1$. For $y$ slightly less than 1 (e.g., $y=0.9$), $\pi y$ is slightly less than $\pi$, so $\sin(\pi y) > 0$. The solution curve moves upward, toward $y=1$. Since solutions on both sides of $y=1$ approach it, the equilibrium is stable, and the slope segments point towards the line.
5. The slope field for $\frac{dy}{dx} = \frac{xy}{x^2 + 1}$ exhibits which symmetry property?
   Answer: Symmetric about the $x$-axis only When analyzing symmetry properties of slope fields, you need to examine how the differential equation behaves under coordinate transformations. Replace $x$ with $-x$, $y$ with $-y$, or both, and see what happens to the expression. For $\frac{dy}{dx} = \frac{xy}{x^2 + 1}$, let's test each symmetry type systematically. To check symmetry about the $x$-axis, replace $y$ with $-y$: $\frac{d(-y)}{dx} = \frac{x(-y)}{x^2 + 1} = -\frac{xy}{x^2 + 1}$. Since $\frac{d(-y)}{dx} = -\frac{dy}{dx}$, this gives us $-\frac{dy}{dx} = -\frac{xy}{x^2 + 1}$, which simplifies back to our original equation. This confirms symmetry about the $x$-axis. Now let's eliminate the other options. For $y$-axis symmetry, replace $x$ with $-x$: $\frac{dy}{dx} = \frac{(-x)y}{(-x)^2 + 1} = \frac{-xy}{x^2 + 1}$. This gives the negative of our original equation, so there's no $y$-axis symmetry. This eliminates options A and B. For origin symmetry, replace both $x$ with $-x$ and $y$ with $-y$: $\frac{d(-y)}{dx} = \frac{(-x)(-y)}{(-x)^2 + 1} = \frac{xy}{x^2 + 1}$. Since $\frac{d(-y)}{dx} = -\frac{dy}{dx}$, we get $-\frac{dy}{dx} = \frac{xy}{x^2 + 1}$, which contradicts our original equation. This eliminates option D. Study tip: For slope field symmetry problems, systematically test each coordinate transformation. Remember that $x$-axis symmetry means the slope field looks the same when flipped over the $x$-axis, which happens when replacing $y$ with $-y$ preserves the differential equation's form.
6. A solution curve for the differential equation $\frac{dy}{dx} = 1 - xy$ passes through the origin $(0,0)$. Which of the following describes the behavior of the solution curve as it leaves the origin?
   Answer: It has an initial slope of 1, and its slope decreases as it moves into the first quadrant. First, find the slope at the origin: $\frac{dy}{dx}$ at $(0,0)$ is $1 - (0)(0) = 1$. So the curve leaves the origin with a slope of 1, moving into the first quadrant where both $x$ and $y$ are positive. In the first quadrant, the term $xy$ is positive. As $x$ and $y$ increase from 0, the product $xy$ increases. Therefore, the slope, given by $1-xy$, will decrease from its initial value of 1.
7. The line segments of the slope field for the differential equation $\frac{dy}{dx} = \frac{xy}{x^2 + y^2 - 9}$ are vertical at certain points in the plane. These points form which of the following shapes?
   Answer: A circle with center (0,0) and radius 3. Vertical line segments correspond to an undefined slope. For a rational function, this occurs when the denominator is equal to zero. Setting the denominator to zero gives $x^2 + y^2 - 9 = 0$, which can be rewritten as $x^2 + y^2 = 9$. This is the equation of a circle centered at the origin with a radius of $\sqrt{9} = 3$.
8. For the differential equation $\frac{dy}{dx} = x^2 - 4$, in which region of the coordinate plane will the slope field show segments with negative slope?
   Answer: $-2 < x < 2$ The slope is negative when $x^2 - 4 < 0$, which occurs when $x^2 < 4$, or $-2 < x < 2$. Note that the slope depends only on $x$, not on $y$. Choice A gives the region where slope is positive. Choices C and D are incorrect because the slope is independent of the $y$-coordinate in this differential equation.
9. For the differential equation $\frac{dy}{dx} = x - \ln(y)$, let $m_1$ be the slope of the solution curve at the point $(2, e)$ and $m_2$ be the slope at the point $(3, e^2)$. Which of the following is true?
   Answer: $m_1 = m_2$ The slope at any point is determined directly by the differential equation, not by a specific solution curve. We calculate the slopes at the given points. At $(2, e)$, the slope is $m_1 = 2 - \ln(e) = 2 - 1 = 1$. At $(3, e^2)$, the slope is $m_2 = 3 - \ln(e^2) = 3 - 2 = 1$. Since both slopes are equal to 1, $m_1 = m_2$. The two points lie on the same isocline.
10. For the differential equation $\frac{dy}{dx} = \cos(x) \cdot y$, at which point would the slope field show the steepest positive slope?
    Answer: $\left(0, 2\right)$ When you encounter a slope field question, you're analyzing how steep the tangent lines are at different points based on the differential equation. The slope at any point $(x,y)$ is given directly by the right side of the equation: $\frac{dy}{dx} = \cos(x) \cdot y$. To find the steepest positive slope, you need to maximize the expression $\cos(x) \cdot y$. Since all answer choices have the same $y$-value of 2, you're really looking for where $\cos(x)$ is largest. The cosine function reaches its maximum value of 1 at $x = 0, 2\pi, 4\pi,$ etc. At point $(0, 2)$, the slope equals $\cos(0) \cdot 2 = 1 \cdot 2 = 2$. This is the steepest positive slope among all the choices, making C correct. Let's check why the other options fall short. At point A $(\pi, 2)$, the slope is $\cos(\pi) \cdot 2 = (-1) \cdot 2 = -2$, which is negative, not positive. At point B $(\frac{\pi}{2}, 2)$, the slope is $\cos(\frac{\pi}{2}) \cdot 2 = 0 \cdot 2 = 0$, which is neither positive nor steep. At point D $(\frac{3\pi}{2}, 2)$, the slope is $\cos(\frac{3\pi}{2}) \cdot 2 = 0 \cdot 2 = 0$, again giving zero slope. Remember: for slope field problems, substitute the coordinates directly into the differential equation. The key is recognizing that maximizing $\cos(x) \cdot y$ means finding where cosine is largest when $y$ values are equal.
11. Consider the differential equation $\frac{dy}{dx} = y - x^2$. Solution curves to this equation can have local extrema. Which of the following statements correctly describes where these extrema can occur?
    Answer: Any local maximum must occur on the parabola $y=x^2$ in the region where $x > 0$. Local extrema occur where $\frac{dy}{dx} = 0$, which means $y - x^2 = 0$, or $y = x^2$. To classify these extrema, we use the second derivative test. $\frac{d^2y}{dx^2} = \frac{d}{dx}(y - x^2) = \frac{dy}{dx} - 2x$. At a point on the parabola $y=x^2$, we have $\frac{dy}{dx}=0$, so $\frac{d^2y}{dx^2} = 0 - 2x = -2x$. For a local maximum, we need $\frac{d^2y}{dx^2} < 0$, which means $-2x < 0$, or $x > 0$. For a local minimum, we need $\frac{d^2y}{dx^2} > 0$, which means $-2x > 0$, or $x < 0$. Thus, local maxima occur on $y=x^2$ for $x>0$.
12. A differential equation of the form $\frac{dy}{dx} = f(x,y)$ is called homogeneous if $f(tx, ty) = f(x,y)$ for any $t \neq 0$. This property implies that the slope $\frac{dy}{dx}$ depends only on the ratio $\frac{y}{x}$. What geometric feature does the slope field of a homogeneous equation possess?
    Answer: Slopes are constant along any line passing through the origin. If the slope $\frac{dy}{dx}$ depends only on the ratio $\frac{y}{x}$, then for any line passing through the origin, given by $y=mx$, the ratio $\frac{y}{x}$ is constant and equal to $m$. Since the slope depends only on this ratio, the slope must be constant all along the line $y=mx$ (excluding the origin). Therefore, the isoclines of a homogeneous equation are lines passing through the origin.
13. For the differential equation $\frac{dy}{dx} = x^2 - y$, in which region of the $xy$-plane are the solution curves concave down?
    Answer: Below the parabola $y = x^2 - 2x$ Concavity is determined by the sign of the second derivative, $\frac{d^2y}{dx^2}$. We differentiate $\frac{dy}{dx}$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(x^2 - y) = 2x - \frac{dy}{dx}$. Substitute the original equation for $\frac{dy}{dx}$: $\frac{d^2y}{dx^2} = 2x - (x^2 - y) = y - x^2 + 2x$. Solution curves are concave down when $\frac{d^2y}{dx^2} < 0$, which means $y - x^2 + 2x < 0$, or $y < x^2 - 2x$. This inequality describes the region below the parabola $y = x^2 - 2x$.
14. The slope field for $\frac{dy}{dx} = f(x,y)$ has positive slopes in quadrants I and III, and negative slopes in quadrants II and IV. Additionally, slopes are zero on the y-axis (for $y \neq 0$) and vertical on the x-axis (for $x \neq 0$). Which equation matches this description?
    Answer: $\frac{dy}{dx} = \frac{x}{y}$ The sign of $\frac{x}{y}$ is positive when $x,y$ have the same sign (quadrants I and III) and negative when they have opposite signs (quadrants II and IV). Slopes are zero when the numerator is zero, so $x=0$ (the y-axis). Slopes are vertical (undefined) when the denominator is zero, so $y=0$ (the x-axis). This matches all the given conditions. Choice A has the correct sign behavior but no vertical slopes. Choice B has the correct sign behavior but the zero/vertical axes are swapped. Choice C does not have the correct sign behavior in all quadrants.
15. The slope field for the differential equation $\frac{dy}{dx} = \frac{x^3}{y^2}$ possesses a certain symmetry. Which of the following describes this symmetry?
    Answer: Symmetry with respect to the y-axis. Let $f(x,y) = \frac{x^3}{y^2}$. A slope field is symmetric with respect to the y-axis if reflecting a point $(x,y)$ and its line segment across the y-axis results in a valid point and segment in the field. This means the slope at $(-x,y)$ must be the negative of the slope at $(x,y)$. So, we test if $f(-x,y) = -f(x,y)$. $f(-x,y) = \frac{(-x)^3}{y^2} = \frac{-x^3}{y^2} = -f(x,y)$. This condition holds, so the field is symmetric with respect to the y-axis. The other symmetries do not hold.
16. The slope field for an autonomous differential equation $\frac{dy}{dx} = f(y)$ indicates that solutions are strictly increasing for $-1 < y < 3$ and strictly decreasing elsewhere. Which of the following is a possible expression for $f(y)$?
    Answer: $f(y) = (y+1)(3-y)$ The information implies that $f(y) > 0$ on the interval $(-1, 3)$ and $f(y) < 0$ for $y < -1$ or $y > 3$. This means $f(y)$ has roots at $y=-1$ and $y=3$. This suggests a quadratic form $k(y+1)(y-3)$. Let's test the sign of $(y+1)(y-3)$. In the interval $(-1,3)$, picking $y=0$, we get $(1)(-3) = -3$, which is negative. Since we need $f(y)$ to be positive in this interval, we must multiply by a negative constant, e.g., $k=-1$. This gives $f(y) = -(y+1)(y-3) = (y+1)(3-y)$. This function is a downward-opening parabola with roots at -1 and 3, so it is positive between the roots and negative elsewhere, matching the description.
17. Consider the slope field for the differential equation $\frac{dy}{dx} = |x| - y$. The set of all points where the line segments are horizontal forms which shape?
    Answer: The line $y=x$ for $x \ge 0$ and $y=-x$ for $x < 0$. Horizontal line segments occur where the slope $\frac{dy}{dx}$ is zero. Setting the equation to zero gives $|x| - y = 0$, which is equivalent to $y = |x|$. The graph of $y = |x|$ is a V-shape composed of two lines: $y=x$ for $x \ge 0$ and $y=-x$ for $x < 0$.
18. The slope field for a certain differential equation has the property that its isoclines are all circles centered at the origin. Which of the following could be the differential equation?
    Answer: $\frac{dy}{dx} = x^2+y^2+1$ Isoclines are curves where the slope is constant. If the isoclines are circles centered at the origin, of the form $x^2+y^2 = r^2$, it means that the slope $\frac{dy}{dx}$ must be constant for a constant value of $x^2+y^2$. This implies that $\frac{dy}{dx}$ must be a function of $x^2+y^2$. Among the choices, only $\frac{dy}{dx} = x^2+y^2+1$ has this form. For choice A, the solution curves are circles, not the isoclines. For choice B, the isoclines are lines through the origin. For choice D, the isoclines are lines of the form $y=-x+C$.
19. Consider the differential equation $\frac{dy}{dx} = \frac{4x^2 + y^2}{2x^2}$. What is the limiting behavior of the slopes of its slope field as $|x| \to \infty$ along the line $y=2x$?
    Answer: The slope approaches 4. To find the limiting behavior of the slope along the line $y=2x$, we substitute $y=2x$ into the expression for $\frac{dy}{dx}$ and evaluate the limit as $|x| \to \infty$. $\frac{dy}{dx} = \frac{4x^2 + (2x)^2}{2x^2} = \frac{4x^2 + 4x^2}{2x^2} = \frac{8x^2}{2x^2} = 4$ for $x \neq 0$. Since the expression simplifies to the constant 4, the slope is 4 at every point on the line $y=2x$ (except the origin). Therefore, the limiting slope is 4.
20. Consider the slope field for $\frac{dy}{dx} = \sin(x+y)$. If you start at the point $(0, \pi)$ and follow the slope field, what is the approximate direction of the initial movement?
    Answer: Horizontally to the right with no vertical component When you encounter a slope field problem, you need to evaluate the differential equation at the given starting point to determine the initial direction of movement. The slope at any point tells you the direction a solution curve will travel from that location. At the starting point $(0, \pi)$, you substitute these coordinates into the differential equation: $\frac{dy}{dx} = \sin(0 + \pi) = \sin(\pi) = 0$. This means the slope of the tangent line at this point is zero, indicating purely horizontal movement to the right with no vertical component. Looking at the wrong answers: Choice A suggests vertical movement, which would require an undefined (infinite) slope, not zero. Choice B indicates downward movement, which would need a negative slope. Choice C suggests upward movement requiring a positive slope. All of these contradict our calculated slope of zero. The key insight is that $\sin(\pi) = 0$ because $\pi$ radians corresponds to 180°, where the sine function equals zero. When the derivative equals zero, the solution curve moves horizontally at that instant—it's neither increasing nor decreasing, just moving rightward along a constant $y$-value. Remember this pattern: always substitute your starting coordinates directly into the given differential equation to find the initial slope. The sign and magnitude of this slope tells you everything about the initial direction. Zero slope means horizontal movement, positive slope means upward movement, and negative slope means downward movement.

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