Reasoning Using Slope Fields
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AP Calculus AB › Reasoning Using Slope Fields
The slope field shown is for $\frac{dy}{dx}=y^2-4$. Which equilibrium solution is stable?
Both $y=-2$ and $y=2$
Neither $y=-2$ nor $y=2$
$y=-2$ only
$y=2$ only
All constant functions are stable
Explanation
This question uses slope field reasoning to identify stable equilibrium solutions. For $dy/dx = y^2 - 4 = (y-2)(y+2)$, equilibrium occurs when $dy/dx = 0$, giving $y = 2$ and $y = -2$. For stability, we check the sign of $dy/dx$ near each equilibrium. Near $y = -2$: slightly above gives $(-1.9)^2 - 4 \approx -0.39 < 0$ (decreasing toward $-2$), slightly below gives $(-2.1)^2 - 4 \approx 0.41 > 0$ (increasing toward $-2$). Near $y = 2$: slightly below gives $(1.9)^2 - 4 \approx -0.39 < 0$ (moving away), slightly above gives $(2.1)^2 - 4 \approx 0.41 > 0$ (moving away). Only $y = -2$ attracts nearby solutions, making it stable. Choice B fails because $y = 2$ is unstable as solutions move away from it. To identify stable equilibria in slope fields, find where $dy/dx = 0$ and check if nearby slopes point toward that equilibrium.
A slope field corresponds to $\frac{dy}{dx}=\frac{1}{1+y^2}$. Which statement about solution curves is true?
They oscillate up and down periodically.
They are decreasing for all $x$.
They have slope between $0$ and $1$ everywhere.
They are constant when $y\neq 0$.
They have vertical tangents when $y=0$.
Explanation
This question involves slope field reasoning for equations with bounded derivatives. For $\frac{dy}{dx} = \frac{1}{1+y^2}$, the denominator $1 + y^2$ is always $\geq 1$ for any real $y$, so the slope is always between 0 and 1. Since $1 + y^2 \geq 1$, we have $\frac{1}{1+y^2} \leq 1$, and since $1 + y^2 > 0$, we have $\frac{dy}{dx} > 0$. Therefore, all solution curves have slopes strictly between 0 and 1 everywhere. Choice A fails because solutions are always increasing (positive slope), never decreasing. When analyzing slope fields, examine the range of possible values for $\frac{dy}{dx}$ to understand the constraints on solution behavior and ensure slopes stay within predictable bounds.
A slope field corresponds to $\frac{dy}{dx}=\frac{x}{1+y^2}$. At $x=0$, solution curves have
slope $1$ for all $y$.
vertical tangents for all $y$.
undefined slope for all $y$.
negative slope for all $y$.
horizontal tangents for all $y$.
Explanation
This question requires slope field reasoning about slopes along vertical lines. For dy/dx = x/(1+y²) at x = 0, we have dy/dx = 0/(1+y²) = 0 for all y values. Since the numerator is zero when x = 0, the slope equals zero regardless of the y-coordinate, making all slope segments horizontal along the vertical line x = 0. The denominator 1+y² is always positive, so the fraction is well-defined and equals zero. Choice A fails because slopes are zero (horizontal), not infinite (vertical), along x = 0. When analyzing slope fields along vertical lines where the differential equation has x in the numerator, substitute the x-value to see how slopes vary with y.
A slope field corresponds to $\frac{dy}{dx}=\frac{2}{1+x^2}-y^2$. At $(0,2)$, the solution is locally
increasing because slope is positive.
undefined because slope is infinite.
increasing because slope is zero.
constant because slope is zero.
decreasing because slope is negative.
Explanation
This question requires slope field reasoning about local solution behavior at specific points. For $\frac{dy}{dx} = \frac{2}{1+x^2} - y^2$ at point (0,2), we evaluate: $\frac{dy}{dx} = \frac{2}{1+0^2} - 2^2 = 2 - 4 = -2 < 0$. A negative slope indicates the solution is locally decreasing at (0,2). The slope magnitude of 2 shows the solution decreases at a moderate rate near this point. Choice A fails because the slope is negative, not positive, indicating decreasing rather than increasing behavior. When analyzing local solution behavior in slope fields, substitute the point's coordinates into the differential equation to determine the slope sign and magnitude at that location.
A slope field corresponds to $\frac{dy}{dx}=e^{-x}$. Which statement best describes the family of solution curves?
Solutions have negative slope for $x>0$.
All solutions increase with slopes that decrease as $x$ increases.
All solutions decrease and are concave down for all $x$.
All solutions have slopes that increase as $x$ increases.
All solutions are constant functions.
Explanation
This question involves slope field reasoning for solutions of equations depending only on x. For dy/dx = e^(-x), the slope depends only on x, not on y, so all solutions are vertical translations of each other. Since e^(-x) > 0 for all x, all solutions are increasing. As x increases, e^(-x) decreases toward 0, so slopes decrease but remain positive, meaning solutions increase at a decreasing rate (concave down). Choice A fails because solutions increase, not decrease, and choice D fails because slopes decrease, not increase. When analyzing slope fields for dy/dx = f(x), remember all solutions have identical slope behavior at each x-value, differing only by vertical shifts.
For $\frac{dy}{dx}=\frac{x}{1+y^2}$, which statement about the solution through $(0,2)$ is true near $x=0$?
It has a horizontal tangent because slope is $0$ at $x=0$
It has a vertical tangent because denominator is $0$
It is decreasing because slopes are negative at $x=0$
It is constant because slope depends only on $y$
It is increasing because slopes are positive at $x=0$
Explanation
This problem tests slope field evaluation at specific points. For dy/dx = x/(1+y²), we evaluate the slope at the point (0,2). Substituting x = 0 into the differential equation gives dy/dx = 0/(1+2²) = 0/5 = 0. A slope of 0 means the solution has a horizontal tangent at this point. The answer "vertical tangent" might seem plausible if one confuses the roles of numerator and denominator, but the denominator 1+y² is never zero. When evaluating slopes from differential equations, substitute the exact coordinates and simplify carefully to determine the tangent line's behavior.
A slope field corresponds to $\frac{dy}{dx}=x(2-y)$. What happens to solutions that start with $y<2$ as $x$ increases?
They remain below $y=2$ and increase toward $y=2$ for $x>0$.
They are constant for all $x$.
They decrease without bound for $x>0$.
They cross $y=2$ repeatedly for $x>0$.
They move away from $y=2$ for all $x>0$.
Explanation
This question involves slope field reasoning to predict long-term solution behavior. For dy/dx = x(2-y), when y < 2 and x > 0, we have (positive)(positive) = positive slope, so solutions increase. As these solutions approach y = 2, the factor (2-y) approaches 0, making the slope approach 0, so the solution levels off approaching y = 2. Since y < 2 throughout this process, solutions never actually reach or cross y = 2. Choice A incorrectly suggests solutions move away from y = 2, but positive slopes when y < 2 drive solutions toward this horizontal asymptote. When reading slope fields, identify equilibrium lines where dy/dx = 0 and determine whether solutions approach or diverge from these lines based on nearby slope signs.
A slope field corresponds to $\frac{dy}{dx}=\frac{y}{y^2+1}$. Which statement about slopes is true for all points?
Slopes depend on $x$ only.
Slopes are undefined when $y=0$.
Slopes are always between $-\tfrac12$ and $\tfrac12$.
Slopes are always greater than $1$.
Slopes are always negative.
Explanation
This question involves slope field reasoning for bounded slope behavior. For dy/dx = y/(y²+1), we can analyze the range of possible slopes. Taking the derivative of f(y) = y/(y²+1), we get f'(y) = (1-y²)/(y²+1)², which equals 0 when y = ±1. Evaluating f(±1) = ±1/2, and noting f(0) = 0, f(y) → 0 as y → ±∞, the function achieves maximum 1/2 at y = 1 and minimum -1/2 at y = -1. Therefore, slopes are always between -1/2 and 1/2. Choice A fails because slopes can be positive when y > 0. To analyze slope bounds in slope fields, find the range of the differential equation by examining its critical points and limiting behavior.
A slope field for $\frac{dy}{dx}=y-x$ is shown. Which statement about the solution through $(0,1)$ is true?
It initially increases and is concave down near $x=0$.
It is constant for all $x$.
It initially decreases and is concave down near $x=0$.
It initially increases and is concave up near $x=0$.
It initially decreases and later stays constant.
Explanation
This question requires using slope field reasoning to analyze solution behavior. For dy/dx = y - x at point (0,1), the slope is 1 - 0 = 1, so the curve is initially increasing. Moving slightly right from (0,1) to small positive x values, the slopes remain positive and even increase since y > x in that region, indicating the curve is concave up. Nearby points show slopes that increase as we move right and up, confirming concave up behavior. Choice A fails because the solution increases rather than decreases initially. To read slope fields effectively, first evaluate the slope at the given point, then examine how slopes change in the immediate neighborhood to determine concavity.
A slope field corresponds to $\frac{dy}{dx}=\ln(1+y)$. Which statement about solutions with $y>-1$ is true?
Solutions with $-1<y<0$ are increasing.
Solutions with $y>0$ are increasing.
All solutions have negative slope.
Slopes depend only on $x$.
All solutions are constant.
Explanation
This question requires slope field reasoning for logarithmic differential equations. For dy/dx = ln(1+y) with y > -1, we need to analyze when ln(1+y) is positive, negative, or zero. Since ln(1+y) > 0 when 1+y > 1, i.e., when y > 0, solutions with y > 0 have positive slope and are increasing. When -1 < y < 0, we have 0 < 1+y < 1, so ln(1+y) < 0, making these solutions decreasing. When y = 0, ln(1) = 0, giving a horizontal tangent. Choice C fails because solutions with -1 < y < 0 are decreasing, not increasing. To analyze slope fields involving logarithmic functions, determine where the logarithmic expression is positive, negative, or zero based on its argument.