Sketching Slope Fields
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AP Calculus BC › Sketching Slope Fields
Which slope field matches $\dfrac{dy}{dx}=y(1-y)$ for a population fraction $y$ over time $x$?
Slopes are zero on $x=0$ and $x=1$; slopes are positive for $0<x<1$ and negative for $x>1$ or $x<0$.
Slopes are zero on $y=0$ and $y=1$; slopes are positive for $0<y<1$ and negative for $y>1$ or $y<0$.
Slopes depend on both variables with zero slopes on $y=x$; slopes change sign across $y=x$.
Slopes are zero only at the origin; slopes are positive in Quadrants I and III and negative in Quadrants II and IV.
Slopes are constant on diagonals $y-x=c$; slopes increase as $c$ increases.
Explanation
Sketching slope fields is a key skill in AP Calculus BC for visualizing solutions to differential equations like dy/dx = y(1 - y). To verify, at y = 0, such as (any x, 0), dy/dx = 0(1-0) = 0, and at y = 1, dy/dx = 1(1-1) = 0, confirming zero slopes. Between 0 < y < 1, at (x, 0.5), dy/dx = 0.5(0.5) = 0.25 > 0, positive, while for y > 1 at (x, 2), dy/dx = 2(-1) = -2 < 0, and for y < 0 at (x, -1), dy/dx = -1(2) = -2 < 0, matching choice A. These points show slopes are independent of x, constant along horizontal lines. A tempting distractor like choice B fails because slopes depend on y, not x; at x=0.5, dy/dx varies with y, not constant along vertical lines. Always verify slope fields by plugging sample points into the differential equation to check consistency with the described features.
Which slope field matches $\dfrac{dy}{dx}=\dfrac{1}{1+y^2}$ for all real $x$ and $y$?
Slopes are zero along $y=0$ and increase with $|y|$.
Slopes depend only on $x$, always positive, and decrease toward $0$ as $|x|$ increases.
Slopes are undefined on $y=0$ and very steep near that line.
Slopes depend only on $y$, always positive, and decrease toward $0$ as $|y|$ increases.
Slopes are negative for $y>0$ and positive for $y<0$.
Explanation
This question tests understanding slope fields for rational functions with restricted ranges. For dy/dx = 1/(1+y²), the denominator 1+y² is always positive and at least 1, so slopes are always positive but never exceed 1. The slope equals 1 when y = 0 (since 1/(1+0²) = 1) and decreases toward 0 as |y| increases (since 1+y² grows without bound). At y = 2, slope = 1/(1+4) = 0.2; at y = -3, slope = 1/(1+9) = 0.1. Choice C incorrectly claims slopes are negative for y > 0, but 1/(1+y²) is always positive. For bounded slope fields, identify maximum and minimum values and where they occur.
Which slope field matches the differential equation $\dfrac{dy}{dx}=xy$?
Slopes depend only on $y$ and are identical across each horizontal row.
Slopes are zero along both axes; positive in Quadrants I and III and negative in Quadrants II and IV.
Slopes are zero along both axes; positive in Quadrants II and IV and negative in Quadrants I and III.
Slopes are zero along $y=x$ only and constant along lines parallel to $y=x$.
Slopes depend only on $x$ and are identical across each vertical column.
Explanation
This question requires sketching the slope field for dy/dx = xy, which depends on both x and y. At (0,0), the slope is 0·0=0; at (1,1), the slope is 1·1=1; at (-1,1), the slope is (-1)·1=-1; at (1,-1), the slope is 1·(-1)=-1. The slopes are zero along both coordinate axes (where x=0 or y=0), positive in Quadrants I and III (where x and y have the same sign), and negative in Quadrants II and IV (where x and y have opposite signs). Choice E incorrectly claims slopes are zero only along y=x, but xy=0 when either x=0 or y=0. For products in differential equations, analyze the sign of each factor to determine the overall sign pattern.
Which slope field matches the differential equation $\dfrac{dy}{dx}=xy$?
Slopes are always negative and become steeper as $x$ increases.
Slopes depend only on $x$; each vertical line has constant slope with zeros on $x=0$.
Slopes depend only on $y$; each horizontal line has constant slope with zeros on $y=1$.
Slopes are zero along both axes; slopes are positive in Quadrants I and III and negative in Quadrants II and IV.
Slopes are zero along $y=x$; slopes are positive when $y<x$ and negative when $y>x$.
Explanation
This question involves sketching slope fields for separable equations with xy dependence. For dy/dx = xy, slopes equal zero when either x = 0 or y = 0, giving horizontal tangents along both coordinate axes. In Quadrant I (x > 0, y > 0) and Quadrant III (x < 0, y < 0), the product xy is positive; in Quadrants II and IV, one factor is negative, making xy negative. Choice B incorrectly places zero slopes along y = x instead of the coordinate axes. To analyze dy/dx = xy, check the sign of the product in each quadrant by considering the signs of x and y separately.
Which slope field matches $\dfrac{dy}{dx}=\cos y$ for a curve $y(x)$ where slopes repeat as $y$ changes?
Along any vertical line $x=c$, all segments share the same slope; slopes vary periodically with $x$.
Slopes are zero along $y=x$; slopes are negative above it and positive below it.
Along any horizontal line $y=c$, all segments share the same slope; slopes vary periodically with $y$.
Slopes are zero along $y=1$ and increase steadily as $y$ increases.
Slopes are zero along $y=0$ and undefined along $x=0$.
Explanation
Sketching slope fields is a key skill in AP Calculus BC for visualizing solutions to differential equations like dy/dx = cos y. To verify, along horizontal y=c, dy/dx = cos c, constant regardless of x. For y=0, cos0=1; y=π/2, cos(π/2)=0; y=π, cosπ=-1; y=3π/2,0; repeating every 2π in y, periodic, matching choice A. These show independence from x, varying with y. A tempting distractor like choice B fails because slopes depend on y, not x; along vertical x=c, dy/dx=cos y varies with y, not constant. Always verify slope fields by plugging sample points into the differential equation to check consistency with the described features.
Which slope field matches the differential equation $\dfrac{dy}{dx}=\dfrac{y}{x}$ for $x\ne 0$?
Slopes are constant on rays $y=mx$; slopes are undefined along $x=0$ and equal $m$ on $y=mx$.
Slopes are constant on circles centered at the origin; slopes are undefined on $x^2+y^2=0$.
Slopes are zero along $y=x$; slopes are positive above $y=x$ and negative below.
Slopes are constant on vertical lines; slopes are undefined along $x=0$ and equal $x$ elsewhere.
Slopes are constant on horizontal lines; slopes are undefined along $y=0$.
Explanation
Sketching slope fields is a key skill in understanding differential equations by visualizing the slopes they dictate at various points. To verify the slope field for dy/dx = y/x (x≠0), note along a ray y=mx, slope=m everywhere on that ray, constant. For example, along y=2x, at (1,2)=2/1=2, at (2,4)=4/2=2; along y=-x, (-1,1)=1/(-1)=-1, (2,-2)=-2/2=-1. Undefined along x=0, and each ray has its own constant slope m. A tempting distractor like choice E fails because along y=x, y/x=1, not zero; zero would require y=0, the x-axis. Always pick test points along proposed lines of zero slope and constant y or x lines to confirm patterns in slope fields.
Which slope field matches the differential equation $\dfrac{dy}{dx}=x-y$ on the $xy$-plane?
Slopes are zero along $y=-x$; along any horizontal line, slopes increase as $x$ increases.
Slopes depend only on $y$; along any horizontal line, all segments have the same slope.
Slopes are constant on vertical lines; slopes are zero along $x=0$ and increase with $y$.
Slopes are zero along $y=x$; along any horizontal line, slopes decrease as $x$ increases.
Slopes are zero along $y=x$; along any horizontal line, slopes increase as $x$ increases.
Explanation
Sketching slope fields is a key skill in understanding differential equations by visualizing the slopes they dictate at various points. To verify the slope field for dy/dx = x - y, check points along y = x, such as (1,1) where the slope is 1-1=0, and (2,2) where it's 2-2=0, confirming zero slopes there. Along a horizontal line like y=1, at x=0 the slope is 0-1=-1, at x=1 it's 1-1=0, and at x=2 it's 2-1=1, showing slopes increase with x. Similarly, along y=0, slopes equal x, increasing positively as x grows. A tempting distractor like choice A fails because it claims slopes decrease as x increases along horizontal lines, but actually they increase since the x term adds positively. Always pick test points along proposed lines of zero slope and constant y or x lines to confirm patterns in slope fields.
Which slope field matches $\dfrac{dy}{dx}=e^{-x}y$ emphasizing how slope changes as $x$ increases?
Slopes depend on both $x$ and $y$; for fixed $y$, slopes decrease in magnitude as $x$ increases; slopes are zero on $y=0$.
Slopes are constant along lines $y=-x$; horizontal on $y=x$.
Slopes depend only on $x$ and approach $0$ as $x$ increases, regardless of $y$.
Slopes depend only on $y$; for fixed $x$, slopes decrease as $x$ increases.
Slopes are zero along $x=0$; slopes increase with $|x|$.
Explanation
This question examines how slopes change with position in exponential decay slope fields. For dy/dx = e^(-x)y, the slope depends on both x and y as a product. When y = 0, the slope is always zero regardless of x, giving a horizontal nullcline. For fixed y ≠ 0, as x increases, e^(-x) decreases toward 0, so the magnitude of the slope decreases. At (0,1), slope = $e^0$ × 1 = 1, while at (2,1), slope = e^(-2) × 1 ≈ 0.135, confirming the decrease. Choice A incorrectly claims slopes depend only on x, ignoring the factor of y. When analyzing slope fields with products, examine how each factor affects the slope independently.
Which slope field matches $\dfrac{dy}{dx}=x-y$ for a solution passing through $(0,0)$?
Slopes are zero along $y=x$; above that line slopes are negative, below it slopes are positive.
Slopes are zero along $x=0$; slopes are positive for $x>0$ and negative for $x<0$.
Slopes are zero along $y=-x$; above that line slopes are positive, below it slopes are negative.
Slopes depend only on $y$, with zero slopes along $y=0$.
All slope segments are horizontal everywhere.
Explanation
This question requires sketching a slope field for a linear differential equation. For dy/dx = x - y, slopes are zero when x = y, which is the line y = x. At point (2,1), the slope is 2-1 = 1 (positive), while at (1,2), the slope is 1-2 = -1 (negative). This confirms that above the line y = x (where y > x), we have x - y < 0, giving negative slopes, and below the line (where y < x), we have x - y > 0, giving positive slopes. Choice D incorrectly identifies y = -x as the zero-slope line, but substituting shows that dy/dx = x - (-x) = 2x, which is only zero when x = 0. When sketching slope fields, always verify your nullclines by substituting back into the differential equation.
Which slope field matches the differential equation $\dfrac{dy}{dx}=y(1-y)$ near the equilibrium solutions?
Slopes are constant everywhere, equal to $1$.
Slopes depend only on $y$, with zero slopes along $y=0$ and $y=1$; positive for $0<y<1$, negative otherwise.
Slopes depend on $x+y$, with zero slopes along the line $y=-x$.
Slopes are zero along $y=x$ and increase with distance from that line.
Slopes depend only on $x$, with zero slopes along $x=0$ and $x=1$.
Explanation
This question tests your ability to sketch slope fields for autonomous differential equations. For dy/dx = y(1-y), the slope depends only on y, not on x, so slopes are constant along horizontal lines. Setting dy/dx = 0 gives y = 0 and y = 1 as equilibrium solutions where slopes are zero. For 0 < y < 1, we have y > 0 and (1-y) > 0, so dy/dx > 0 (positive slopes). For y < 0 or y > 1, the product y(1-y) is negative, giving negative slopes. Choice A incorrectly suggests slopes depend on x, when they actually depend only on y. To verify slope fields, always check: where are slopes zero, what determines the slope value, and what are the signs in different regions.