Sketching Slope Fields
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AP Calculus AB › Sketching Slope Fields
Which slope field corresponds to $\dfrac{dy}{dx}=\dfrac{y+1}{y-1}$ (with $y\neq 1$)?
Slopes depend only on $x$; undefined along $x=1$; slopes are $0$ along $x=-1$.
Slopes are periodic in $y$ with zeros at $y=k\pi$.
Slopes are $0$ along $y=1$ and undefined along $y=-1$.
Slopes depend only on $y$; undefined along $y=1$; slopes are $0$ along $y=-1$.
Slopes are $0$ along $x=0$ and undefined along $y=0$.
Explanation
This question asks for the slope field matching dy/dx = (y + 1)/(y - 1) with y ≠ 1. Slopes depend only on y since the equation involves only y. The function is undefined along y = 1. Slopes are zero when y + 1 = 0, so y = -1. For y > 1, both numerator and denominator are positive, creating positive slopes. For -1 < y < 1, the numerator is positive while denominator is negative, creating negative slopes. For y < -1, both factors are negative, making slopes positive. Choice C incorrectly reverses which line has zero slopes versus undefined slopes. For rational functions (y + a)/(y + b), find where numerator and denominator separately equal zero.
Which slope field corresponds to $\dfrac{dy}{dx}=e^{-x}$?
Slopes depend only on $y$; always positive; steeper for smaller $y$.
Slopes are $0$ along $y=e^{-x}$ and change sign across the curve.
Slopes are $0$ along $x=0$ and negative for $x>0$.
Slopes alternate sign periodically as $x$ increases.
Slopes depend only on $x$; always positive; steeper for smaller $x$ and flatten as $x$ increases.
Explanation
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 $\dfrac{dy}{dx}=\sin x$?
Slopes are constant along diagonals $y-x=\text{constant}$.
Slopes depend only on $y$ and repeat periodically as $y$ increases.
Slopes are always positive but approach $0$ as $x$ increases.
Slopes depend only on $x$ and alternate sign periodically in vertical columns; $0$ at multiples of $\pi$.
Slopes are $0$ along $y=\sin x$; positive above the curve and negative below.
Explanation
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 corresponds to $\dfrac{dy}{dx}=\dfrac{x}{y}$ (with $y\neq 0$)?
Slopes depend only on $x$ and are undefined at $x=0$.
Slopes are undefined along $y=0$; slopes are $0$ along $x=0$; sign matches $x/y$.
Slopes depend only on $y$ and are undefined at $y=0$.
Slopes are undefined along $x=0$ and $0$ along $y=0$.
Slopes are constant along rays $y=mx$ and undefined at $x=0$.
Explanation
This question requires sketching the slope field for dy/dx = x/y with y ≠ 0. The function is undefined along y = 0 (x-axis), creating a horizontal asymptote in the slope field. Slopes are zero when x = 0 (y-axis). The sign of slopes matches the sign of x/y: positive when x and y have the same sign (quadrants I and III), negative when they have opposite signs (quadrants II and IV). This creates a pattern where slopes become very steep as y approaches zero. Choice A incorrectly reverses the roles of x and y in determining zeros versus undefined behavior. For rational functions x/y, the x-axis is undefined while the y-axis gives zero slopes.
Which slope field matches $\dfrac{dy}{dx}=(y-1)(y+2)$?
Slopes are $0$ along $x=-2$ and $x=1$; positive between them.
Slopes depend only on $x$; $0$ at $x=\pm 1$.
Slopes are $0$ along $y=0$ and $y=3$; positive between.
Slopes are $0$ along $y=-2$ and $y=1$; positive for $y>1$ and $y<-2$, negative for $-2<y<1$.
Slopes are $0$ along $y=x$ and change sign across it.
Explanation
This question involves sketching the slope field for dy/dx = (y - 1)(y + 2), where slopes depend only on y. All points in the same horizontal row share identical slopes. Slopes are zero when either factor equals zero: y = 1 or y = -2. For y > 1, both factors are positive, creating positive slopes. For -2 < y < 1, the first factor is negative while the second is positive, creating negative slopes. For y < -2, both factors are negative, making their product positive. Choice D incorrectly identifies y = 0 and y = 3 as equilibrium points rather than y = -2 and y = 1. For factored expressions, find where each factor equals zero, then analyze the sign of the product in each interval.
Which slope field corresponds to $\dfrac{dy}{dx}=\tan x$ (where defined)?
Slopes depend only on $y$ and have vertical asymptotes at $y=\frac{\pi}{2}+k\pi$.
Slopes are $0$ along $y=x$ and undefined along $y=-x$.
Slopes depend only on $x$; always between $-1$ and $1$.
Slopes depend only on $x$; $0$ at $x=k\pi$; very steep near $x=\frac{\pi}{2}+k\pi$.
All slopes are positive and constant.
Explanation
This question asks for the slope field matching dy/dx = tan x where defined. Slopes depend only on x since tan x involves only x. The tangent function has vertical asymptotes where cos x = 0, so at x = π/2 + kπ. Slopes are zero where sin x = 0, so at x = kπ. Between consecutive asymptotes, slopes range from -∞ to +∞, creating a periodic pattern. The steep behavior near x = π/2 + kπ distinguishes this from other trigonometric slope fields. Choice A incorrectly suggests slopes depend on y, but tan x involves only the x-variable. For tangent functions, identify both the zeros (where sine equals zero) and the vertical asymptotes (where cosine equals zero).
Which slope field corresponds to $\dfrac{dy}{dx}=\dfrac{y-2}{x+1}$ (where defined)?
Slopes are $0$ along $y=2$; undefined along $x=-1$; sign changes across both lines.
Slopes depend only on $x$; undefined at $x=-1$; positive for $x>-1$.
Slopes are $0$ along $y=-x$ and undefined along $y=x$.
Slopes are $0$ along $x=-1$; undefined along $y=2$.
Slopes depend only on $y$; $0$ along $y=2$; positive above.
Explanation
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 $\dfrac{dy}{dx}=x^3$?
Slopes depend only on $x$; negative for $x<0$, positive for $x>0$, and steeper as $|x|$ increases.
Slopes are $0$ along $y=x^3$ and change sign across the curve.
Slopes depend only on $x$; always nonnegative; symmetric about the $y$-axis.
Slopes depend only on $y$; negative for $y<0$ and positive for $y>0$.
All slopes are constant and negative.
Explanation
This question involves sketching the slope field for dy/dx = x³, where slopes depend only on x. All points in the same vertical column share identical slopes since the equation involves only x. Slopes are zero when x³ = 0, so only at x = 0 (the y-axis). For x > 0, slopes are positive and increase rapidly as x increases. For x < 0, slopes are negative and become increasingly steep as x decreases. The cubic function creates steeper slopes than quadratic functions, with slopes growing without bound as |x| increases. Choice B incorrectly suggests slopes are always nonnegative, but x³ < 0 when x < 0. For odd-powered functions, slopes change sign across the axis and grow rapidly with distance.
Which slope field matches $\dfrac{dy}{dx}=\dfrac{1}{1+y^2}$?
Slopes depend only on $x$; always positive, largest near $x=0$.
Slopes are $0$ along $y=x$ and change sign across it.
Slopes are positive above $y=0$ and negative below $y=0$.
Slopes are $0$ on $y=0$ and negative elsewhere.
Slopes depend only on $y$; always positive, largest near $y=0$, flattening as $|y|$ increases.
Explanation
This question involves sketching the slope field for dy/dx = 1/(1 + y²), where slopes depend only on y. All points in the same horizontal row share identical slopes. The denominator 1 + y² is always positive, so slopes are always positive regardless of y-value. Slopes are largest when the denominator is smallest, which occurs at y = 0 where the slope equals 1. As |y| increases, the denominator grows, causing slopes to flatten toward zero but never reaching zero. Choice D incorrectly suggests slopes can be negative, but the fraction is always positive. For rational functions with positive denominators, slopes maintain the same sign throughout and are largest where the denominator is minimized.
Which slope field corresponds to $\dfrac{dy}{dx}=y-x^2$?
Slopes are $0$ along $y=x^2$; positive above the parabola and negative below.
Slopes depend only on $y$ and are $0$ along $y=0$.
Slopes are $0$ along $y=-x$; positive above and negative below.
Slopes are $0$ along $x=y^2$; positive to the right and negative to the left.
Slopes depend only on $x$ and are always nonnegative.
Explanation
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.