Sketching Slope Fields - AP Calculus AB

Which slope field matches $\dfrac{dy}{dx}=\dfrac{x-y}{x+y}$ (where defined)?
Answer: Slopes are $0$ along $y=x$ ; undefined along $y=-x$ ; pattern symmetric about the origin. This question involves sketching the slope field for dy/dx = (x - y)/(x + y) where defined. The function is undefined when x + y = 0, so along the line y = -x. Slopes are zero when x - y = 0, so along the line y = x. These two perpendicular lines divide the plane into four regions with different slope behaviors. The pattern is symmetric about the origin due to the structure of the rational function. Choice B incorrectly reverses which line has zero slopes versus undefined slopes. For rational functions involving linear combinations in numerator and denominator, set each to zero separately to find the critical lines.
Which slope field corresponds to $\dfrac{dy}{dx}=y-x^2$ ?
Answer: Slopes are $0$ along $y=x^2$ ; positive above the parabola and negative below. This question requires sketching the slope field for dy/dx = y - x², where slopes depend on both variables. Slopes are zero when y - x² = 0, so y = x² (along the parabola). Above the parabola where y > x², slopes are positive since y - x² > 0. Below the parabola where y < x², slopes are negative since y - x² < 0. The parabolic curve y = x² serves as the boundary between positive and negative slope regions. Choice E incorrectly suggests slopes are zero along y = -x, but the equation requires y = x² for zero slope. When analyzing dy/dx = g(x,y), set the right-hand side equal to zero to find curves of horizontal tangents, then test signs on either side.
Which slope field corresponds to $\dfrac{dy}{dx}=y(2-y)$ in a population model with carrying capacity $2$ ?
Answer: Slopes are $0$ on the horizontal lines $y=0$ and $y=2$ ; positive for $0<y<2$ , negative for $y>2$ . This question involves sketching a slope field for dy/dx = y(2 - y), a logistic differential equation with carrying capacity 2. The slope depends only on y, so all points in the same horizontal row share identical slopes. Slopes are zero when y(2 - y) = 0, which occurs at y = 0 and y = 2. For 0 < y < 2, both factors are positive, making slopes positive. For y > 2, we have y > 0 but (2 - y) < 0, making slopes negative. Choice A incorrectly states slopes depend on x, but this equation shows y-dependence only. To analyze logistic models, find the equilibrium points where growth rate equals zero, then check the sign between equilibria.
Which slope field corresponds to $\dfrac{dy}{dx}=\dfrac{x^2-y^2}{2}$ ?
Answer: Slopes are $0$ along $y=\pm x$ ; positive where $|x|>|y|$ and negative where $|y|>|x|$ . This question requires sketching the slope field for dy/dx = (x² - y²)/2, where slopes depend on both variables. Slopes are zero when x² - y² = 0, so x² = y² or y = ±x. These two lines y = x and y = -x divide the plane into four regions. Slopes are positive when x² > y² (where |x| > |y|) and negative when x² < y² (where |y| > |x|). The diagonal lines create a pattern where slopes are positive in the regions farther from the y-axis and negative in regions closer to the y-axis. Choice B incorrectly suggests a parabolic zero-slope curve rather than the linear boundaries y = ±x. For expressions like x² - y², factor as (x-y)(x+y) or note where |x| = |y| to find zero-slope lines.
Which slope field matches $\dfrac{dy}{dx}=\dfrac{x+y}{x-y}$ (where defined)?
Answer: Slopes are undefined along $y=x$ ; slopes are $0$ along $y=-x$ ; pattern changes across both diagonals. This question involves sketching the slope field for dy/dx = (x + y)/(x - y) where defined. The function is undefined when x - y = 0, so along the line y = x. Slopes are zero when x + y = 0, so along the line y = -x. These two perpendicular lines create four regions with different slope behaviors. The line y = -x provides horizontal tangents while y = x represents a vertical asymptote where slopes become infinite. Choice B incorrectly reverses which line has zero slopes versus undefined slopes. For rational functions of the form (ax + by)/(cx + dy), set the numerator to zero for zero-slope lines and the denominator to zero for undefined-slope lines.
Which slope field matches $\dfrac{dy}{dx}=x^2$ for all $(x,y)$ ?
Answer: Slopes depend only on $x$ ; all slopes are nonnegative, $0$ on $x=0$ , steeper as $|x|$ increases. This question involves sketching the slope field for dy/dx = x², where slopes depend only on x-coordinate. At any point (x, y), the slope equals x², which is always nonnegative. Slopes are zero only when x = 0 (along the y-axis). For x ≠ 0, slopes are positive and increase as |x| increases, creating steeper segments farther from the y-axis. All points in the same vertical column share identical slopes since the equation depends only on x. Choice A incorrectly suggests slopes can be negative, but x² ≥ 0 always. When the differential equation depends on only one variable, look for where that expression equals zero and note that slopes remain constant along lines perpendicular to that variable's axis.
Which slope field matches $\dfrac{dy}{dx}=y\cos x$ ?
Answer: Slopes are $0$ along $y=0$ and along vertical lines where $\cos x=0$ ; sign depends on both $y$ and $x$ . This question involves sketching the slope field for dy/dx = y cos x, where slopes depend on both variables. Slopes are zero when either factor equals zero: y = 0 (x-axis) or cos x = 0 (vertical lines at x = π/2 + kπ). The sign depends on both factors: slopes are positive when y and cos x have the same sign, negative when they have opposite signs. Above the x-axis (y > 0), slopes follow the sign of cos x, while below the x-axis (y < 0), slopes have opposite sign to cos x. Choice D incorrectly suggests slopes depend only on y, missing the crucial x-dependence through cos x. For products involving trigonometric functions, identify where each factor equals zero and analyze the combined sign pattern.
Which slope field matches $\dfrac{dy}{dx}=y^2-1$ ?
Answer: Slopes are $0$ along $y=1$ and $y=-1$ ; negative for $-1<y<1$ , positive for $|y|>1$ . This question involves sketching the slope field for dy/dx = y² - 1, where slopes depend only on y. All points in the same horizontal row share identical slopes. Slopes are zero when y² - 1 = 0, so y = ±1. For |y| < 1 (between y = -1 and y = 1), we have y² < 1, making y² - 1 < 0 (negative slopes). For |y| > 1 (above y = 1 or below y = -1), we have y² > 1, making y² - 1 > 0 (positive slopes). The horizontal lines y = ±1 separate regions of different slope signs. Choice A incorrectly reverses the sign pattern between the equilibrium lines. For quadratic expressions in y, factor to find zeros, then test the sign in each interval.
Which slope field corresponds to $\dfrac{dy}{dx}=\sin x$ ?
Answer: Slopes depend only on $x$ and alternate sign periodically in vertical columns; $0$ at multiples of $\pi$ . This question requires sketching a slope field for dy/dx = sin x, where slopes depend only on the x-coordinate. All points in the same vertical column share identical slopes since the equation involves only x. Slopes are zero when sin x = 0, which occurs at x = kπ for integer k. Slopes alternate between positive and negative as x increases, following the periodic pattern of sine. Between zeros, slopes reach maximum magnitude of 1 at x = π/2 + kπ. Choice A incorrectly suggests slopes depend on y, but sin x involves only the x-variable. For trigonometric differential equations, identify the zeros and extrema of the trig function to understand the periodic slope pattern.
Which slope field matches $\dfrac{dy}{dx}=x^2-y$ ?
Answer: Slopes are $0$ along $y=x^2$ ; positive below the parabola and negative above. This question involves sketching the slope field for dy/dx = x² - y, where slopes depend on both variables. Slopes are zero when x² - y = 0, so y = x² (along the parabola). Below the parabola where y < x², slopes are positive since x² - y > 0. Above the parabola where y > x², slopes are negative since x² - y < 0. The parabolic curve y = x² serves as the boundary between positive and negative slope regions. Choice B incorrectly reverses the sign pattern above and below the parabola. When analyzing dy/dx = f(x) - y, set f(x) = y to find the zero-slope curve, then test signs above and below.
Which slope field matches $\dfrac{dy}{dx}=\dfrac{y}{x}$ (with $x\neq 0$ )?
Answer: Along each ray from the origin ( $y=mx$ ), slopes are constant; slopes are $0$ on the $x$ -axis and undefined at $x=0$ . This question involves sketching the slope field for dy/dx = y/x (with x ≠ 0), a homogeneous differential equation. Along any ray from the origin where y = mx (constant slope m), we have dy/dx = mx/x = m, so slopes are constant along each ray. Slopes are zero along the x-axis (y = 0) and undefined at x = 0. The slope equals the slope of the ray itself, creating a radial pattern. Points farther from the origin along the same ray have identical slopes. Choice B incorrectly suggests slopes depend only on y, but the ratio y/x creates the ray-dependent pattern. For homogeneous equations of the form dy/dx = f(y/x), slopes remain constant along rays from the origin.
A cooling object satisfies $\dfrac{dy}{dx}=-(y-5)$ . Which slope field matches this differential equation?
Answer: Slopes are $0$ along $y=5$ ; negative above $y=5$ and positive below; same within each horizontal row. This question asks for the slope field matching dy/dx = -(y - 5), a cooling model where temperature y approaches ambient temperature 5. The slope depends only on y, creating identical slopes across each horizontal row. Slopes are zero when -(y - 5) = 0, so y = 5. When y > 5, we have (y - 5) > 0, making -(y - 5) < 0 (negative slopes). When y < 5, we have (y - 5) < 0, making -(y - 5) > 0 (positive slopes). Choice D incorrectly suggests equilibria at both y = 0 and y = 5, but only y = 5 gives zero slope. For cooling/heating problems, identify the ambient temperature where slope equals zero, then check signs above and below that value.
Which slope field corresponds to $\dfrac{dy}{dx}=\dfrac{y-2}{x+1}$ (where defined)?
Answer: Slopes are $0$ along $y=2$ ; undefined along $x=-1$ ; sign changes across both lines. This question asks for the slope field matching dy/dx = (y - 2)/(x + 1) where defined. The function is undefined when x + 1 = 0, so along x = -1. Slopes are zero when y - 2 = 0, so along y = 2. These two perpendicular lines divide the plane into regions with different slope characteristics. The line y = 2 provides horizontal tangents while x = -1 represents a vertical asymptote where slopes become infinite. The sign of slopes depends on the signs of both numerator and denominator factors. Choice B incorrectly reverses which line has zero slopes versus undefined slopes. For rational functions, identify the zeros of numerator and denominator separately to understand the complete slope behavior.
Which slope field matches $\dfrac{dy}{dx}=\dfrac{x}{1+y^2}$ ?
Answer: Slopes are $0$ along $x=0$ ; sign matches $x$ ; slopes flatten as $|y|$ increases for fixed $x$ . This question involves sketching the slope field for dy/dx = x/(1 + y²), where slopes depend on both variables but in a specific pattern. Slopes are zero when x = 0 (along the y-axis). The sign of slopes matches the sign of x: positive for x > 0 and negative for x < 0. For any fixed x-value, slopes flatten as |y| increases since the denominator 1 + y² grows. The y-dependence creates a flattening effect without changing the sign, which is determined solely by x. Choice A incorrectly suggests slopes depend only on y, missing the crucial x-dependence for sign determination. When analyzing rational expressions, identify which variable determines the sign and which affects the magnitude.
Which slope field corresponds to $\dfrac{dy}{dx}=\dfrac{y}{1+x^2}$ ?
Answer: Slopes are $0$ along $y=0$ ; sign matches $y$ ; for fixed $y$ , slopes flatten as $|x|$ increases. This question requires sketching the slope field for dy/dx = y/(1 + x²), where slopes depend on both variables. Slopes are zero when y = 0 (along the x-axis). The sign of slopes matches the sign of y: positive for y > 0 and negative for y < 0. For any fixed y-value, slopes flatten as |x| increases since the denominator 1 + x² grows. The x-dependence creates a flattening effect without changing the sign, which is determined solely by y. Choice B incorrectly reverses the roles of x and y in determining zeros and sign patterns. When analyzing products or quotients, identify which factor determines the sign and which affects the magnitude.
Which slope field corresponds to $\dfrac{dy}{dx}=e^{-x}$ ?
Answer: Slopes depend only on $x$ ; always positive; steeper for smaller $x$ and flatten as $x$ increases. This question requires sketching the slope field for dy/dx = $e^{-x}$ , where slopes depend only on x. All points in the same vertical column share identical slopes. The exponential function $e^{-x}$ is always positive, so slopes are always positive throughout the plane. Slopes are largest when x is smallest (moving leftward) and approach zero as x increases (moving rightward). At x = 0, the slope equals 1, and slopes decay exponentially for positive x while growing exponentially for negative x. Choice B incorrectly suggests slopes depend on y, but $e^{-x}$ involves only the x-variable. For exponential functions, note they never reach zero but can approach zero or infinity depending on the sign of the exponent.
Which slope field corresponds to $\dfrac{dy}{dx}=\sin y$ ?
Answer: Slopes depend only on $y$ ; $0$ at $y=k\pi$ ; periodic in horizontal rows. This question requires sketching the slope field for dy/dx = sin y, where slopes depend only on y. All points in the same horizontal row share identical slopes since the equation involves only y. Slopes are zero when sin y = 0, which occurs at y = kπ for integer k. Slopes are positive when sin y > 0 and negative when sin y < 0, following the periodic pattern of sine. The slope pattern repeats every 2π units in the y-direction, creating horizontal stripes of alternating slope signs. Choice A incorrectly suggests slopes depend on x, but sin y involves only the y-variable. For trigonometric equations in y, the slope pattern creates periodic horizontal bands.
Which slope field corresponds to $\dfrac{dy}{dx}=xy$ ?
Answer: Slopes are $0$ along both axes; positive in quadrants I and III, negative in II and IV; steeper farther from axes. This question asks for the slope field matching dy/dx = xy, where slopes depend on both x and y coordinates. Slopes are zero when xy = 0, which occurs along both coordinate axes (x = 0 or y = 0). In quadrant I (x > 0, y > 0), both factors are positive, creating positive slopes. In quadrant III (x < 0, y < 0), both factors are negative, making their product positive. In quadrants II and IV, the factors have opposite signs, yielding negative slopes. Slopes become steeper as points move farther from the axes since |xy| increases. Choice C incorrectly claims slopes are zero along y = x rather than along the axes. For products like xy, find where each factor equals zero and analyze the sign in each quadrant.
Which slope field corresponds to $\dfrac{dy}{dx}=\dfrac{1}{x}$ (with $x\neq 0$ )?
Answer: Slopes depend only on $x$ ; positive for $x>0$ , negative for $x<0$ , very steep near $x=0$ . This question asks for the slope field matching dy/dx = 1/x (with x ≠ 0), where slopes depend only on x. The function is undefined at x = 0, creating a vertical asymptote. For x > 0, slopes are positive with 1/x > 0. For x < 0, slopes are negative with 1/x < 0. Slopes become very steep (large magnitude) as x approaches zero from either side. As |x| increases, slopes flatten toward zero but never reach zero. Choice B incorrectly suggests slopes depend on y, but 1/x involves only the x-variable. For rational functions with vertical asymptotes, identify where the denominator equals zero and analyze the sign and behavior on either side.
Which slope field matches the differential equation $\dfrac{dy}{dx}=x-y$ for all $(x,y)$ ?
Answer: Slopes are $0$ along the line $y=x$ ; slopes positive below $y=x$ and negative above. This question requires sketching a slope field for the differential equation dy/dx = x - y. The slope at any point (x, y) equals x - y, so slopes are zero when x = y (along the line y = x). When x > y (below the line y = x), slopes are positive since x - y > 0. When x < y (above the line y = x), slopes are negative since x - y < 0. Choice A incorrectly suggests slopes are zero along y = -x rather than y = x. The key strategy is to identify where the right-hand side equals zero to find lines of horizontal tangents, then determine the sign pattern on either side.

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