Reasoning Using Slope Fields - AP Calculus BC

Lines of increasing slope parallel to the $y$-axis. Slope depends only on $x$, creating vertical patterns.

Slopes that can be separated into functions of $x$ and $y$. Variables separate into independent $f(x)$ and $g(y)$ functions.

Slopes decrease as $y$ increases. Negative coefficient creates decreasing exponential behavior.

Slope is $5$. At $(0,2)$: $y' = e^0 + 2^2 = 1 + 4 = 5$.

A linear solution. Uniform slope produces straight-line solutions.

Slope is $2$. At $(0,-2)$: $y' = 0^2 - (-2) = 2$.

Slope is $0$. At $(0,3)$: $y' = \frac{0}{2} = 0$.

Slope is $1$. At $(1,1)$: $y' = \frac{1}{1} = 1$.

Slope is $0$. At $(0,0.5)$: $y' = 2(0)(0.5) = 0$.

Slope is $\frac{1}{2}$. At $(2,1)$: $y' = \frac{1}{2}$.

Slope is $-3$. At $(3,3)$: $y' = 3 - 2(3) = -3$.

Slope is $2$. At $(2,4)$: $y' = 3(2) - 4 = 2$.

All lines have slope $1$. Constant derivative creates uniform slope throughout field.

Slope is $2$. Substitute $(1,1)$: $y' = 1 + 1 = 2$.

$y = \frac{k}{2}x^2 + C$. Integration of $kx$ with arbitrary constant $C$.

Slope is $0$. At $(0,0)$: $y' = 0 + 0 = 0$.

Slope is $0$. When $y = 0$, derivative equals zero.

Points where $x^2 = y^2$. When $x^2 = y^2$, the derivative equals zero.

Exponential growth. Positive feedback creates exponentially increasing curves.

Curve that follows the direction of the slopes. Path tangent to slope field segments at every point.

Slope is $1$. At $(1,1)$: $y' = \frac{1^2}{1} = 1$.

Slopes oscillate between positive and negative. Tangent function creates periodic vertical asymptotes.

Slope is $0$. When $y = 0$, derivative equals zero.

Slope is $4$. At $(1,2)$: $y' = 2(1) + 2 = 4$.

Undefined. Division by zero at origin makes slope undefined.

The behavior of differential equation solutions. Shows qualitative solution behavior without solving analytically.

Slope is $3$. At $(1,2)$: $y' = 1 + 2 = 3$.

Slope is $0$. At $(0,-1)$: $y' = 0^3 = 0$.

Slope is $1$. At $(0,1)$: $y' = \frac{1}{1} = 1$.