Calculus Study Guide: Derivatives, Applications, and Integrals

Derivatives and Their Applications

Finding Tangent and Normal Lines

The tangent line to a curve at a given point represents the instantaneous rate of change of the function at that point, while the normal line is perpendicular to the tangent.

• Tangent Line: The equation of the tangent line to a curve at x = a is given by: $y = f'(a)(x - a) + f(a)$

• Normal Line: The normal line at x = a has a slope of $-1/f'(a)$ (provided $f'(a) \neq 0$): $y = -\frac{1}{f'(a)}(x - a) + f(a)$

• Example: For the curve $4x^3 = y^2(4 - y^2)$ (cissoid of Diocles), to find the tangent and normal lines at $x = 2$, first solve for $y$ at $x = 2$, then compute $\frac{dy}{dx}$ using implicit differentiation, and substitute into the formulas above.

Related Rates Problems

Related rates involve finding the rate at which one quantity changes with respect to another, often using implicit differentiation.

• Key Steps:

  1. Identify all variables and their rates of change.

  2. Write an equation relating the variables.

  3. Differentiating both sides with respect to time $t$.

  4. Substitute known values and solve for the desired rate.

• Example: An observer stands 200 meters from a launch site. A balloon rises vertically at 4 m/s. How fast is the angle $\theta$ (between the ground and the observer's line of sight to the balloon) increasing 30 seconds after launch?

  • Let $y$ be the height of the balloon, $x = 200$ m (constant), $y = 4t$.

  • $\tan \theta = \frac{y}{x}$, so $\theta = \arctan\left(\frac{y}{x}\right)$.

  • Differentiate: $\frac{d\theta}{dt} = \frac{1}{1 + (y/x)^2} \cdot \frac{1}{x} \cdot \frac{dy}{dt}$

  • At $t = 30$ s, $y = 120$ m. Substitute values to find $\frac{d\theta}{dt}$.

Curve Sketching and Analysis

Increasing/Decreasing Intervals and Local Extrema

To determine where a function is increasing or decreasing, and to find local maxima and minima, use the first derivative test.

• First Derivative: $f'(x)$ indicates where the function increases ($f'(x) > 0$) or decreases ($f'(x) < 0$).

• Critical Points: Points where $f'(x) = 0$ or $f'(x)$ is undefined.

• Local Extrema: Use the sign of $f'(x)$ around critical points to determine if they are maxima or minima.

• Example: For $f(x) = x^4 + 4x^3$:

  • $f'(x) = 4x^3 + 12x^2$

  • Set $f'(x) = 0$ to find critical points.

Concavity and Inflection Points

Concavity describes the direction a curve bends. Inflection points are where the concavity changes.

• Second Derivative: $f''(x) > 0$ means the function is concave up; $f''(x) < 0$ means concave down.

• Inflection Points: Points where $f''(x) = 0$ and the sign of $f''(x)$ changes.

• Example: For $f(x) = x^4 + 4x^3$:

  • $f''(x) = 12x^2 + 24x$

  • Set $f''(x) = 0$ to find possible inflection points.

Optimization Problems

Maximizing Area with Constraints

Optimization involves finding the maximum or minimum value of a function subject to constraints.

• Example 1: A rectangular bird sanctuary with one side along a riverbank and 12 km of fencing for the other three sides. Find dimensions to maximize area.

  • Let $x$ = length parallel to river, $y$ = width perpendicular.

  • Constraint: $x + 2y = 12$

  • Area: $A = x \cdot y$

  • Express $A$ in terms of one variable, differentiate, set derivative to zero to find maximum.

• Example 2: Given area $A = 8$ km$^2$, minimize fencing for three sides.

  • Constraint: $x \cdot y = 8$

  • Fencing: $x + 2y$

  • Express fencing in terms of one variable, use calculus to find minimum.

Integration and Antiderivatives

Basic Integration Techniques

Integration is the reverse process of differentiation. The antiderivative of a function $f(x)$ is a function $F(x)$ such that $F'(x) = f(x)$.

• Power Rule: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, $n \neq -1$

• Common Integrals:

  • $\int \cos(ax) dx = \frac{1}{a} \sin(ax) + C$

  • $\int \sec^2(ax) dx = \frac{1}{a} \tan(ax) + C$

• Examples:

  • $\int (3x^3 + 1) dx = \frac{3}{4}x^4 + x + C$

  • $\int (2x + 1)^2 dx$ (expand and integrate term by term)

  • $\int 2 \sec^2(x) dx = 2 \tan(x) + C$

Finding Antiderivatives

To find the antiderivative, apply the appropriate rule or substitution.

• Example: $f(x) = 2\sqrt{x} + 6\sec^2(3x)$

  • $\int 2\sqrt{x} dx = \int 2x^{1/2} dx = \frac{4}{3}x^{3/2} + C$

  • $\int 6\sec^2(3x) dx = 2 \tan(3x) + C$

  • So, $\int f(x) dx = \frac{4}{3}x^{3/2} + 2\tan(3x) + C$

Summary Table: Key Calculus Concepts

Concept	Definition	Key Formula	Example
Derivative	Instantaneous rate of change	$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$	$f(x) = x^2 \Rightarrow f'(x) = 2x$
Critical Point	$f'(x) = 0$ or undefined	Set $f'(x) = 0$	$f(x) = x^2 \Rightarrow x = 0$
Inflection Point	Concavity changes	Set $f''(x) = 0$ and check sign change	$f(x) = x^3 \Rightarrow x = 0$
Integral	Area under the curve	$\int f(x) dx$	$\int x dx = \frac{1}{2}x^2 + C$