Sketching Slope Fields
Help Questions
AP Calculus AB › Sketching Slope Fields
Which slope field corresponds to $\dfrac{dy}{dx}=\dfrac{1}{x}$ (with $x\neq 0$)?
Slopes are $0$ along $x=0$ and increase with $|x|$.
Slopes depend only on $x$; positive for $x>0$, negative for $x<0$, very steep near $x=0$.
Slopes depend only on $y$; positive for $y>0$, negative for $y<0$.
Slopes are $0$ along $y=x$ and symmetric about that line.
All slopes are positive and constant.
Explanation
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 corresponds to $\dfrac{dy}{dx}=xy$?
Slopes are $0$ along $y=x$; positive above and negative below.
Slopes depend only on $x$; same in each vertical column.
Slopes are $0$ along $y=0$ and $y=1$; positive between them.
Slopes are $0$ along both axes; positive in quadrants I and III, negative in II and IV; steeper farther from axes.
Slopes are always negative and get steeper as $|x|$ increases.
Explanation
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 $\dfrac{dy}{dx}=\sqrt{|y|}$?
Slopes are positive for $y>0$ and negative for $y<0$.
Slopes are $0$ along $y=\pm 1$ and negative between them.
Slopes depend only on $x$; nonnegative everywhere; $0$ along $x=0$.
Slopes are undefined along $y=0$.
Slopes depend only on $y$; nonnegative everywhere; $0$ along $y=0$; steeper as $|y|$ increases.
Explanation
This question asks for the slope field matching dy/dx = √|y|, where slopes depend only on y. All points in the same horizontal row share identical slopes. Slopes are zero only when |y| = 0, so only along y = 0. The square root function ensures all slopes are nonnegative. For y ≠ 0, slopes are positive and increase as |y| increases, since √|y| grows with distance from the y-axis. The absolute value makes the slope pattern symmetric about the x-axis. Choice C incorrectly suggests slopes can be negative, but √|y| ≥ 0 always. For expressions involving absolute values and square roots, note the nonnegativity and symmetry properties that arise.
Which slope field matches $\dfrac{dy}{dx}=\dfrac{x+y}{1+x^2}$?
Slopes are $0$ along $y=-x$; sign changes across that line; for fixed $(x+y)$, slopes flatten as $|x|$ increases.
Slopes are $0$ along $y=x$; sign changes across that line; for fixed $(x-y)$, slopes flatten as $|x|$ increases.
Slopes are undefined along $x=0$.
Slopes depend only on $x$ and are always positive.
Slopes depend only on $y$ and flatten as $|y|$ increases.
Explanation
This question involves sketching the slope field for dy/dx = (x + y)/(1 + x²), where slopes depend on both variables. Slopes are zero when x + y = 0, so along the line y = -x. Above this line where y > -x, slopes are positive since x + y > 0. Below this line where y < -x, slopes are negative since x + y < 0. For any fixed value of (x + y), slopes flatten as |x| increases since the denominator 1 + x² grows. Choice B incorrectly identifies y = x rather than y = -x as the zero-slope line. For rational functions with numerators of the form ax + by, set the numerator to zero to find the zero-slope line.
Which slope field corresponds to $\dfrac{dy}{dx}=\sin y$?
Slopes are always positive and flatten as $|y|$ increases.
Slopes are $0$ along $y=x$ and positive above.
Slopes are $0$ along $y=\sin x$ and change sign across the curve.
Slopes depend only on $x$; $0$ at $x=k\pi$; periodic in vertical columns.
Slopes depend only on $y$; $0$ at $y=k\pi$; periodic in horizontal rows.
Explanation
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 $\dfrac{dy}{dx}=\dfrac{x^2-y^2}{2}$?
Slopes are undefined along $y=\pm x$.
Slopes are $0$ along $y=\pm x$; positive where $|x|>|y|$ and negative where $|y|>|x|$.
Slopes depend only on $y$; always nonpositive.
Slopes are $0$ along $y=x^2$; positive above, negative below.
Slopes depend only on $x$; always nonnegative.
Explanation
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 the differential equation $\dfrac{dy}{dx}=x-y$ for all $(x,y)$?
Slopes depend only on $y$; equal slopes across each horizontal row.
Slopes are $0$ along the line $y=-x$; slopes increase moving upward.
Slopes depend only on $x$; equal slopes down each vertical column.
Slopes are $0$ along the line $y=x$; slopes positive below $y=x$ and negative above.
All slopes are positive; segments tilt upward everywhere.
Explanation
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.
Which slope field corresponds to $\dfrac{dy}{dx}=\dfrac{y}{y^2+4}$?
Slopes depend only on $y$; $0$ along $y=0$; sign matches $y$; slopes flatten as $|y|$ increases.
Slopes are always positive and increase with $|y|$.
Slopes are $0$ along $y=\pm 2$ and undefined at $y=0$.
Slopes depend only on $x$; $0$ along $x=0$; sign matches $x$.
Slopes are $0$ along $y=x$ and undefined along $y=-x$.
Explanation
This question requires sketching the slope field for dy/dx = y/(y² + 4), where slopes depend only on y. All points in the same horizontal row share identical slopes. 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. The denominator y² + 4 is always positive (≥ 4), so it doesn't affect the sign but does affect magnitude. Slopes are largest near y = 0 and flatten as |y| increases. Choice B incorrectly suggests slopes depend on x, but the equation involves only y. For rational functions y/g(y) where g(y) > 0, slopes have the same sign as y and are maximized where the denominator is minimized.
Which slope field matches $\dfrac{dy}{dx}=y\cos x$?
Slopes depend only on $y$ and are periodic; never zero.
Slopes are $0$ along $y=0$ and along vertical lines where $\cos x=0$; sign depends on both $y$ and $x$.
Slopes depend only on $x$ and are periodic; never zero.
Slopes are $0$ along $y=x$ and change sign across it.
Slopes are $0$ along $x=0$ and along horizontal lines where $\cos y=0$.
Explanation
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 $\dfrac{dy}{dx}=y-x$?
Slopes are $0$ along $y=-x$; positive above and negative below.
Slopes are $0$ along $y=x$; positive above $y=x$ and negative below.
Slopes depend only on $x$; $0$ along $x=0$.
Slopes depend only on $y$; $0$ along $y=0$.
All slopes are positive in quadrant IV.
Explanation
This question involves sketching the slope field for dy/dx = y - x, where slopes depend on both variables. Slopes are zero when y - x = 0, so along the line y = x. Above this line where y > x, slopes are positive since y - x > 0. Below this line where y < x, slopes are negative since y - x < 0. The line y = x serves as the boundary between positive and negative slope regions, with slopes becoming steeper as points move farther from this line. Choice A incorrectly identifies y = -x as the zero-slope line, but the equation requires y = x. For linear combinations like y - x, set the expression equal to zero to find the dividing line, then analyze signs on either side.