MAT 131 Calculus I Practice Exam 2 Rubric

Vocabulary and key concepts from MAT 131 Calculus I Practice Exam 2 Rubric (Chapters 3 and 4), covering integration rules, Riemann sums, displacement, differential equations, and applications.

Anti-Power Rule

The integration rule used to evaluate each term in an indefinite integral, such as integrating terms to obtain 2x^3 - 8\root\relax\blank\to\null\relax\blank\to\null\relax\blank\to\null 2x^3 - 8\root\relax\blank\to\null 2x^3 - 8\root\relax\blank 2x^3 - 8\root\relax 2x^3 - 8\root 2x^3 - 8\root{x} + 3x + C.

Average Rate of Change of Electricity Demand

The average rate of change from 6 a.m. to noon measured in megawatts per hour, expressed as $\frac{P(12) - P(6)}{6}$.

Proportional Acceleration Toward Equilibrium

The second-order differential equation $y'' = -ky$, where acceleration is directly proportional to displacement and oriented back toward equilibrium.

Slope Condition for $\frac{dy}{dx} = -\frac{x}{y}$

A differential equation whose direction field features slopes that are zero along the $y$-axis (except at the origin), undefined along the $x$-axis, and strictly negative in the first quadrant.

Logarithmic Differentiation

The technique of taking the natural logarithm of both sides of y = x^{\root{x}} to obtain \ln(y) = \root{x}\thinspace\text{ln}(x), differentiating to get \frac{1}{y}\thinspace\frac{dy}{dx} = \frac{\text{ln}(x) + 2}{2\root{x}}, and solving for \frac{dy}{dx} = \frac{x^{\root{x}}(\text{ln}(x) + 2)}{2\root{x}}.

Elasticity of Demand ($E$)

The ratio $E = \frac{dq/q}{dp/p} = \frac{dq}{dp} \times \frac{p}{q}$; when $\frac{dq}{dp} = q\thinspace\text{ln}(0.9)$ at $p = 5$, $E = 5\thinspace\text{ln}(0.9) \thickapprox -0.527$, indicating demand decreases by approximately $0.53\%$.

Upper Sum ($U$) for Bar Mass

An upper estimate of total mass calculated using six subintervals of width $0.5\thinspace\text{meter}$ with subinterval maxima $4$, $7$, $8$, $8$, $7$, and $5\thinspace\text{kg/m}$, yielding $U = 19.5\thinspace\text{kg}$.

Lower Sum ($L$) for Bar Mass

A lower estimate of total mass calculated using six subintervals of width $0.5\thinspace\text{meter}$ with subinterval minima $3$, $4$, $7$, $7$, $5$, and $4\thinspace\text{kg/m}$, yielding $L = 15\thinspace\text{kg}$.

Net Change of Linear Density

The application of the Net Change Theorem to the density derivative: $\int_0^3 r'(x)\,dx = r(3) - r(0) = 4 - 3 = 1\thinspace\text{kg/m}$, indicating linear density is $1\thinspace\text{kg/m}$ greater at the $x = 3$ end than at the $x = 0$ end.

Displacement via Signed Integral

The net change in position computed as $\int_0^{12} v(t)\,dt = 12 - 2\times\text{π} \thickapprox 5.72\thinspace\text{meters}$, meaning the cyclist ends approximately $5.72\thinspace\text{meters}$ east of the initial starting point.

Farthest East Position

The maximum eastward position of the cyclist found by evaluating $x(t) = 5 + \int_0^t v(s)\,ds$, which yields $x(4) = 13\thinspace\text{m}$ at $t = 4\thinspace\text{seconds}$ compared to $x(12) = 17 - 2\times\text{π} \thickapprox 10.72\thinspace\text{m}$.

Total Distance Traveled

The sum of the magnitudes of the enclosed areas under the velocity graph, calculated as $\int_0^{12} |v(t)|\,dt = 12 + 2\times\text{π} \thickapprox 18.28\thinspace\text{meters}$.

Greenhouse Temperature Derivative ($T'(t)$)

The instantaneous rate of change of temperature found by the Chain Rule as $T'(t) = \frac{\text{π}}{3}\thinspace\text{cos}\thinspace\text{π}t/12$, which equals T'(9) = -\frac{\text{π}\root{2}}{6} \thickapprox -0.740^\text{o}\text{C/hour}.

Average Temperature from Midnight to Noon

The mean temperature over a 12-hour interval given by $\frac{1}{12}\int_0^{12} T(t)\,dt = 20 + \frac{8}{\text{π}} \thickapprox 22.55^\text{o}\text{C}$.

Salt Mixing Differential Equation

The rate-of-change balance equation $S' = 0.75 - \frac{S}{200}$, formed by taking the salt-entry rate ($0.75\thinspace\text{kg/min}$) minus the salt-exit rate ($\frac{S}{200}\thinspace\text{kg/min}$) for a constant volume of $5000\thinspace\text{liters}$.

Salt Amount Function ($S(t)$)

The particular solution to the salt mixing differential equation with initial amount $S(0) = 20\thinspace\text{kg}$, expressed as $S(t) = 150 - 130e^{-t/200}$.