Sketching Slope Fields
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AP Calculus BC › Sketching Slope Fields
Which slope field matches $\dfrac{dy}{dx}=\cos y$ for a curve $y(x)$ where slopes repeat as $y$ changes?
Along any horizontal line $y=c$, all segments share the same slope; slopes vary periodically with $y$.
Along any vertical line $x=c$, all segments share the same slope; slopes vary periodically with $x$.
Slopes are zero along $y=0$ and undefined along $x=0$.
Slopes are zero along $y=x$; slopes are negative above it and positive below it.
Slopes are zero along $y=1$ and increase steadily as $y$ increases.
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 $\dfrac{dy}{dx}=y(1-y)$ for a population fraction $y$ over time $x$?
Slopes are zero only at the origin; slopes are positive in Quadrants I and III and negative in Quadrants II and IV.
Slopes depend on both variables with zero slopes on $y=x$; slopes change sign across $y=x$.
Slopes are constant on diagonals $y-x=c$; slopes increase as $c$ increases.
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 are zero on $x=0$ and $x=1$; slopes are positive for $0<x<1$ and negative for $x>1$ or $x<0$.
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 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 vertical lines; slopes are undefined along $x=0$ and equal $x$ elsewhere.
Slopes are constant on horizontal lines; slopes are undefined along $y=0$.
Slopes are zero along $y=x$; slopes are positive above $y=x$ and negative below.
Slopes are constant on circles centered at the origin; slopes are undefined on $x^2+y^2=0$.
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 decrease as $x$ increases.
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 zero along $y=x$; along any horizontal line, slopes increase as $x$ increases.
Slopes are constant on vertical lines; slopes are zero along $x=0$ and increase with $y$.
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 only on $y$; for fixed $x$, slopes decrease as $x$ increases.
Slopes depend only on $x$ and approach $0$ as $x$ increases, regardless of $y$.
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 zero along $x=0$; slopes increase with $|x|$.
Slopes are constant along lines $y=-x$; horizontal on $y=x$.
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 $x=0$; slopes are positive for $x>0$ and negative for $x<0$.
Slopes depend only on $y$, with zero slopes along $y=0$.
Slopes are zero along $y=x$; above that line slopes are negative, below it slopes are positive.
Slopes are zero along $y=-x$; above that line slopes are positive, below it slopes are negative.
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 depend only on $x$, with zero slopes along $x=0$ and $x=1$.
Slopes are constant everywhere, equal to $1$.
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 $y$, with zero slopes along $y=0$ and $y=1$; positive for $0<y<1$, negative otherwise.
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.
Which slope field matches the differential equation $\dfrac{dy}{dx}=\sin x$ on $-\pi\le x\le \pi$?
Slopes are constant and equal to $\sin(1)$ everywhere.
Slopes depend only on $x$, with zero slopes at $x=-\pi,0,\pi$ and positive on $(0,\pi)$, negative on $(-\pi,0)$.
Slopes depend only on $y$, repeating periodically as $y$ changes.
Slopes are zero along $x=0$ only, and increase with $|y|$.
Slopes are zero along $y=x$ and $y=-x$.
Explanation
This question tests recognizing slope fields for trigonometric differential equations. For dy/dx = sin x, the slope depends only on x, not on y, so slopes are constant along vertical lines. The sine function equals zero at x = -π, 0, and π, giving horizontal slope segments at these x-values. Between x = 0 and x = π, sin x > 0, so slopes are positive; between x = -π and x = 0, sin x < 0, so slopes are negative. Choice A incorrectly suggests slopes depend on y, but sin x is independent of y. To sketch trigonometric slope fields, recall the key features of the trig function: zeros, sign changes, and periodicity.
Which slope field matches $\dfrac{dy}{dx}=y^2-x$ based on where the slope segments are horizontal?
Horizontal segments occur along $y=\pm\sqrt{x}$ for $x\ge 0$; slopes increase as $y^2-x$ increases.
Horizontal segments occur along $y=x$ only; slopes depend only on $x-y$.
No horizontal segments occur anywhere; slopes are never zero.
Horizontal segments occur along $x=\pm\sqrt{y}$ for $y\ge 0$; slopes depend only on $x$.
Horizontal segments occur along $y=0$ and $y=1$ only; slopes depend only on $y$.
Explanation
This question focuses on finding nullclines (where slopes are zero) in slope fields. For dy/dx = y² - x, setting the slope to zero gives y² = x, which means x = y² or equivalently y = ±√x for x ≥ 0. These are two parabolic curves where slope segments are horizontal. Above these curves (where y² > x), slopes are positive; below them (where y² < x), slopes are negative. Choice C incorrectly identifies y = x as the only nullcline, but substituting gives dy/dx = x² - x = x(x-1), which is zero at x = 0 and x = 1, not along the entire line. When finding nullclines, solve the equation dy/dx = 0 algebraically and verify by checking specific points.
Which slope field matches $\dfrac{dy}{dx}=\sin(x)$ for $-\pi\le x\le \pi$?
Slopes are zero along $y=x$; slopes are positive above that line and negative below it.
Slopes depend only on $x$; slopes are zero at $x=-\pi,0,\pi$ and positive on $(0,\pi)$.
Slopes are constant on diagonal lines $y-x=c$; slopes are zero along $y=x$.
Slopes depend only on $x$; slopes are zero at $x=-\pi,0,\pi$ and negative on $(0,\pi)$.
Slopes depend only on $y$; along each horizontal line, all segments match.
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 = sin(x), note it depends only on x, so vertical lines have constant slopes, like at x=0 where sin(0)=0 everywhere. At x=π/2, sin(π/2)=1>0, constant for all y; at x=3π/2, sin(3π/2)=-1<0, also constant. Zeros occur at x=-π,0,π as sin vanishes there, and positive on (0,π) where sin(x)>0. A tempting distractor like choice C fails because it claims negative slopes on (0,π), but sin(x) is actually positive in that interval. Always pick test points along proposed lines of zero slope and constant y or x lines to confirm patterns in slope fields.