Sketching Slope Fields - AP Calculus BC

Which slope field matches $\dfrac{dy}{dx}=e^x$ , where slopes are always positive and increase as $x$ increases?
Answer: Slopes form vertical bands, always positive, and get steeper to the right. This question requires sketching the slope field for dy/dx = $e^x$ , where slopes depend only on x. Since $e^x$ > 0 for all x, slopes are always positive. As x increases, $e^x$ increases exponentially: at x=0, slope = $e^0$ = 1; at x=1, slope = $e^1$ ≈ 2.72; at x=-1, slope = $e^{-1}$ ≈ 0.37. This creates vertical bands of constant slope that get steeper moving right. Option C incorrectly claims slopes are negative for x>0, but $e^x$ is always positive. Option A suggests horizontal bands, which would require dependence on y alone. For exponential functions in slope fields, remember they're always positive (for $e^x$ ) and change rapidly, creating dramatic differences between left and right regions.
Which slope field matches $\dfrac{dy}{dx}=\dfrac{1}{y}$ for a process where slope depends on current level $y$ ?
Answer: Slopes are undefined on $y=0$ ; slopes are positive for $y>0$ , negative for $y<0$ , and flatten as $|y|$ increases. Sketching slope fields is a key skill in AP Calculus BC for visualizing solutions to differential equations like dy/dx = 1/y. To verify, on y = 0, dy/dx is undefined due to division by zero, as stated. For y > 0, at (x,1), dy/dx = 1/1 = 1 > 0, and at (x,2), dy/dx = 1/2 = 0.5 > 0 but flatter; for y < 0, at (x,-1), dy/dx = -1 < 0, and at (x,-2), dy/dx = -0.5 < 0 but less steep, matching choice A. These points show slopes flatten as |y| increases and are independent of x. A tempting distractor like choice C fails because on y=0 it's undefined, not zero, and slopes flatten, not steepen, with |y|. 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 $\dfrac{dy}{dx}=\sin x$ ?
Answer: Slopes depend only on $x$ and repeat periodically; each vertical column has the same slope. This question tests sketching slope fields for $dy/dx = \sin(x)$ , which depends only on $x$ . At $x=0$ , the slope is $\sin(0)=0$ ; at $x=\pi/2$ , the slope is $\sin(\pi/2)=1$ ; at $x=\pi$ , the slope is $\sin(\pi)=0$ ; at $x=3\pi/2$ , the slope is $\sin(3\pi/2)=-1$ . Since the derivative depends only on $x$ , all points in a vertical column (same $x$ -value) have identical slopes, and the pattern repeats with period $2\pi$ . Choice B incorrectly suggests slopes depend on $y$ and are constant in horizontal rows, but $\sin(x)$ clearly depends only on $x$ . For derivatives that depend on a single variable, identify which variable it is and remember that slopes are constant along lines perpendicular to that variable's axis.
Which slope field matches the differential equation $\dfrac{dy}{dx}=y(1-y)$ ?
Answer: Slopes are zero along $y=0$ and $y=1$ ; positive for $0<y<1$ and negative for $y<0$ or $y>1$ . This question requires sketching the slope field for dy/dx = y(1-y), which depends only on y. At y=0, the slope is 0(1-0)=0; at y=1, the slope is 1(1-1)=0; at y=0.5, the slope is 0.5(0.5)=0.25 (positive); at y=2, the slope is 2(-1)=-2 (negative). The zero slopes occur along the horizontal lines y=0 and y=1, with positive slopes between them (0<y<1) and negative slopes outside (y<0 or y>1). Choice B incorrectly claims the zeros occur along vertical lines x=0 and x=1, but the equation shows dy/dx depends only on y, not x. When the derivative depends on only one variable, all segments in lines perpendicular to that variable's axis have identical slopes.
Which slope field matches $\dfrac{dy}{dx}=y(1-y)$ for a population fraction $y$ over time $x$ ?
Answer: Slopes are zero on $y=0$ and $y=1$ ; slopes are positive for $0<y<1$ and negative for $y>1$ or $y<0$ . 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 $\dfrac{dy}{dx}=\dfrac{y}{x}$ , with equal slopes on rays from the origin and undefined on $x=0$ ?
Answer: Slopes are constant on rays $y=mx$ and are undefined along $x=0$ . This question tests sketching slope fields for dy/dx = y/x, which is undefined along x=0 (the y-axis). Along any ray from the origin with equation y = mx (where m is constant), we have dy/dx = mx/x = m, so slopes are constant along rays. For example, along y = 2x, all slopes equal 2; along y = -x, all slopes equal -1. This creates a radial pattern emanating from the origin. Option A incorrectly suggests slopes are constant on circles, but y/x changes as you move around a circle. Option D claims slopes are constant on horizontal lines, but y/x varies as x changes along such lines. For rational functions like y/x, look for self-similar patterns and lines where the expression simplifies.
Which slope field matches $\dfrac{dy}{dx}=\cos y$ , where slopes are constant along horizontal lines and repeat as $y$ changes?
Answer: Slopes are constant along horizontal lines and vary periodically with $y$ . This question involves sketching the slope field for dy/dx = cos y, where slopes depend only on y. Since cosine is periodic with period 2π, the slope pattern repeats vertically as y increases. Along any horizontal line (constant y), all slopes equal cos y: at y=0, slopes = cos(0) = 1; at y=π/2, slopes = cos(π/2) = 0; at y=π, slopes = cos(π) = -1. This creates horizontal bands of constant slope that repeat every 2π units vertically. Option B incorrectly suggests vertical bands, which would require dependence on x alone. Option D mentions undefined slopes, but cos y is defined for all y. When slope fields depend on trigonometric functions of one variable, expect periodic patterns perpendicular to that variable's axis.
Which slope field matches $\dfrac{dy}{dx}=xy$ for a system where slope is zero on both coordinate axes?
Answer: Slopes are zero on $x=0$ and $y=0$ , positive in Quadrants I and III, negative in II and IV. This question requires sketching the slope field for dy/dx = xy. Slopes are zero when xy = 0, which occurs on both coordinate axes (x=0 or y=0). In Quadrant I where x>0 and y>0, slopes are positive; in Quadrant II where x<0 and y>0, slopes are negative; in Quadrant III where x<0 and y<0, slopes are positive; in Quadrant IV where x>0 and y<0, slopes are negative. This creates a pattern where slopes are positive in Quadrants I and III, negative in II and IV. Option A incorrectly states slopes are positive everywhere except y=0, missing the sign changes. For products like xy, analyze the sign in each quadrant by considering the signs of both factors.
Which slope field matches the differential equation $\dfrac{dy}{dx}=xy$ ?
Answer: Slopes are zero along both axes; slopes are positive in Quadrants I and III and negative in Quadrants II and IV. 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 the differential equation $\dfrac{dy}{dx}=\dfrac{1}{y}$ ?
Answer: Slopes depend only on $y$ ; positive for $y>0$ , negative for $y<0$ , and very steep near $y=0$ . This question tests sketching slope fields for dy/dx = 1/y, which depends only on y. At y=1, the slope is 1/1=1; at y=2, the slope is 1/2=0.5; at y=-1, the slope is 1/(-1)=-1; as y approaches 0, the slopes become very steep (approaching ±∞). The slopes are positive for y>0, negative for y<0, and undefined at y=0, with steeper slopes near y=0. Choice B incorrectly suggests the pattern depends on x, but 1/y clearly depends only on y. For rational functions in slope fields, identify where the derivative is undefined and note that slopes become very steep near these points.
Which slope field matches the differential equation $\dfrac{dy}{dx}=e^{-y}$ ?
Answer: Slopes depend only on $y$ and are always positive; slopes decrease as $y$ increases. This question involves sketching the slope field for dy/dx = $e^{-y}$ , which depends only on y. Since $e^{-y}$ is always positive for all real y, all slopes are positive. At y=0, the slope is $e^0$ =1; at y=1, the slope is $e^{-1}$ ≈0.368; at y=-1, the slope is $e^1$ ≈2.718. As y increases, $e^{-y}$ decreases, so slopes decrease as y increases but remain positive. Choice C incorrectly claims slopes are zero along y=0 and negative for y>0, but $e^{-y}$ is never zero or negative. For exponential functions in slope fields, remember that $e^{anything}$ is always positive, determining the sign pattern.
Which slope field matches the differential equation $\dfrac{dy}{dx}=xy$ ?
Answer: Slopes are zero along both axes; positive in Quadrants I and III and negative in Quadrants II and IV. 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 corresponds to $\dfrac{dy}{dx}=x^2$ for position $y$ changing with time $x$ ?
Answer: Slopes depend only on $x$ ; all slopes are nonnegative and get steeper as $|x|$ increases. Sketching slope fields is a key skill in AP Calculus BC for visualizing solutions to differential equations like dy/dx = x². To verify, slopes depend only on x, so at x=0, dy/dx=0, nonnegative. At x=1, dy/dx=1, and at x=2, dy/dx=4, steeper; at x=-1, dy/dx=1, and x=-2, dy/dx=4, symmetric and steeper with |x|, matching choice A. These points confirm constant slopes along vertical lines, always nonnegative. A tempting distractor like choice B fails because slopes depend on x, not y; at fixed y, dy/dx varies with x, not constant along horizontal 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 $\dfrac{dy}{dx}=x-y$ for a moving particle's path $y(x)$ ?
Answer: Slopes are zero along the line $y=x$ ; above it slopes are negative and below it slopes are positive. Sketching slope fields is a key skill in AP Calculus BC for visualizing solutions to differential equations like dy/dx = x - y. To verify, pick a point on y = x, such as (1,1), where dy/dx = 1 - 1 = 0, confirming zero slope. Above the line, at (1,2), dy/dx = 1 - 2 = -1, a negative slope, and below at (1,0), dy/dx = 1 - 0 = 1, a positive slope. These calculations match the description in choice A across multiple points. A tempting distractor like choice C fails because for y = -x, at (1,-1), dy/dx = 1 - (-1) = 2 ≠ 0, and signs don't match the stated pattern. 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 $\dfrac{dy}{dx}=\dfrac{y}{x}$ ?
Answer: Slopes are constant along rays $y=kx$ ; slopes are positive when $x$ and $y$ share a sign. This question tests sketching slope fields for dy/dx = y/x, which has special properties. Along any ray y=kx from the origin, the slope is (kx)/x = k, which is constant. At (1,2), the slope is 2/1=2; at (2,4), the slope is 4/2=2; at (-1,1), the slope is 1/(-1)=-1. The slopes are positive when x and y have the same sign (Quadrants I and III) and negative when they have opposite signs (Quadrants II and IV). Choice B incorrectly suggests slopes are constant along circles, but y/x is constant along rays, not circles. For homogeneous differential equations like dy/dx = y/x, slopes are constant along rays from the origin.
Which slope field matches the differential equation $\dfrac{dy}{dx}=e^y$ ?
Answer: Slopes depend only on $y$ ; all slopes are positive and increase rapidly as $y$ increases. This question tests slope fields for exponential functions of y. Since dy/dx = $e^y$ depends only on y, horizontal lines have constant slope. The exponential function $e^y$ is always positive and increases rapidly as y increases, with $e^0$ = 1 at y = 0. Choice C incorrectly suggests slopes are zero at y = 0 and negative for y < 0, but $e^y$ > 0 for all real y. For exponential slope fields, remember that $e^y$ grows without bound as y increases and approaches 0 (but remains positive) as y approaches negative infinity.
Which slope field corresponds to $\dfrac{dy}{dx}=xy$ for a quantity $y$ with growth rate proportional to $x$ and $y$ ?
Answer: Slopes are zero on both axes; slopes are positive in Quadrants I and III and negative in Quadrants II and IV. Sketching slope fields is a key skill in AP Calculus BC for visualizing solutions to differential equations like dy/dx = x y. To verify, on axes, if x=0, dy/dx=0 regardless of y; if y=0, dy/dx=0 regardless of x, zero on both. In quadrant I (x>0,y>0), dy/dx>0; quadrant III (x<0,y<0), >0; quadrant II (x<0,y>0), <0; quadrant IV (x>0,y<0), <0, matching choice A. These sign patterns align with the product x y. A tempting distractor like choice D fails because slopes depend on both x and y; along horizontal y=c>0, dy/dx = c x varies with x, 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 $\dfrac{dy}{dx}=y^2$ ?
Answer: Slopes are zero on $y=0$ ; slopes are nonnegative everywhere and increase as $|y|$ increases. This question involves sketching slope fields for autonomous equations with quadratic dependence. For dy/dx = y², slopes depend only on y and are always nonnegative since squares are non-negative. Slopes equal zero only when y = 0, and they increase as |y| increases because larger |y| values produce larger y². Choice C incorrectly suggests negative slopes for y > 0, contradicting the fact that y² ≥ 0 for all real y. When sketching fields for dy/dx = f(y), remember that horizontal lines have constant slope, and the sign of f(y) determines the slope direction.
Which slope field matches the differential equation $\dfrac{dy}{dx}=1-y^2$ ?
Answer: Slopes are zero along $y=1$ and $y=-1$ ; positive for $-1<y<1$ and negative for $|y|>1$ . This question requires sketching the slope field for dy/dx = 1 - y², which depends only on y. Setting 1-y²=0 gives y²=1, so y=±1 are the zero-slope lines. At y=0, the slope is 1-0=1 (positive); at y=0.5, the slope is 1-0.25=0.75 (positive); at y=2, the slope is 1-4=-3 (negative). The slopes are positive for -1<y<1 (where y² < 1) and negative for |y|>1 (where y² > 1). Choice C incorrectly reverses the sign pattern, claiming slopes are negative between y=±1, but 1-y² is positive when y² < 1. To verify sign patterns, test specific points in each region defined by the zero-slope curves.
Which slope field matches $\dfrac{dy}{dx}=x-y$ , where slope is zero along a line through the origin?
Answer: Slopes are zero along $y=x$ and increase moving right or down. This question involves sketching the slope field for dy/dx = x - y. To find where slopes are zero, set x - y = 0, giving y = x, a line through the origin with slope 1. Testing points: at (2,1), slope = 2-1 = 1 (positive); at (1,2), slope = 1-2 = -1 (negative). Moving right from the line y=x increases x more than y, making slopes positive; moving down decreases y, also making slopes positive. Option A incorrectly identifies y=-x as the zero-slope line, but that would require x-(-x)=0, giving 2x=0 only at x=0. When sketching slope fields with expressions like x±y, first find the zero-slope line, then determine how slopes change as you move away from it.

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