8 min readβ’january 24, 2023

Sumi Vora

Kashvi Panjolia

There are many different kinds of functions in math because not everything in the world exists on a plane with two variables. So far, everything we have been doing has been on the **Cartesian** plane: β^2. This is also known as the xy-plane. However, some functions that model the world around us are better graphed using other types of planes, which we will explore in this unit. This unit makes up **11-12%** of the AP Calculus BC Exam.

As you are reading through this guide, pay special attention to the **formulas** mentioned. This unit is very formula-heavy, and ideally you should have all these formulas memorized, but some of them you can derive on the exam.

A parametric equation is typically written in the form:

x = f(t)
y = g(t)

where x and y are the coordinates of a point on the curve, and t represents time. By changing the value of t, we can trace out the entire curve defined by the parametric equations. In a parametric function, both the x and y variables are **dependent** **variables,** and time is the **independent variable.**

To find the **derivative** of a parametric function, we need to find the derivative of x(t) and y(t) and set y'(t) over x'(t). When we do this, the dt's cancel out and we are left with the derivative dy/dx.

As with equations in the Cartesian plane, we can take the **second derivative** of a parametric function. The process for finding the second derivative is a bit different than the process you are used to. We use the chain rule after finding the first derivative to arrive at this equation for the second derivative of a parametric function:

Notice how inside the parentheses, the formula states we need to find dy/dx, not dy/dt. This means that to find the second derivative, you must first find the first derivative with respect to x, then take the derivative of the first derivative (usually using the quotient rule), then set all of that over the first derivative of x(t). As you can see, there are quite a few steps involved, but with some practice, you will master second derivatives in no time.

The **arc length** of a function is a measure of the distance along a curve defined by the function. More specifically, it is the length of the curve between two points. Remember that for Cartesian equations, the formula for the arc length of a curve was:

The same logic still applies to parametric equations, but the formula looks a bit different since x and y are both dependent variables. This is the formula for the arc length of a parametric equation:

In this formula, you are still squaring the derivative of the equation, but since there are two dependent variables, we square the derivatives of both variables. Remember to still take the square root of the sum of the two derivatives and take the integral across your interval. Instead of being "the integral from x=a to x=b," we now say "the integral from t=a to t=b" because t is now the independent variable, instead of x like we are used to.

A **vector-valued function** is a function that maps a real number to a vector in a vector space. It is written in the form

r(t) = <f(t), g(t)> or r(t) = f(t)i + g(t)j

where f(t) and g(t) are real-valued functions and i and j are the unit vectors in the x and y direction respectively. This can be represented geometrically as a point in space moving in the xy-plane.

For parametric equations, vector-valued functions are used to represent the **position**, **velocity** and **acceleration** of an object moving in space. All these functions are related to each other in the way that velocity is the first derivative of position and acceleration is the second derivative of position.

In order to differentiate a vector-valued function, you simply differentiate each of its **components** individually. For example, to go from this vector-valued function for position:

s(t) = <3x+2, ln(x+9)>

to the vector-valued function for velocity, which is the derivative of the position function, you would simply take the derivative of the x-component and the y-component, but you don't need to combine them:

v(t) = s'(t) = <3, 1/(x+9)>

Remember, the derivative rule for the natural log of a function is u'/u where u is the expression inside the parentheses.

For example, let's take the integral of a velocity function that is vector-valued to find the displacement of the object:

v(t) = <2x, 3x^2>

To take this integral, we will integrate the x and y components separately, like this:

s(t) = β«v(t) = <β« 2x dx, β« 3x^2 dx>

Answer: s(t) = β«v(t) = <x^2, x^3>

To solve a motion problem using parametric equations, we first need to identify the parametric equations that describe the position of the object. Once we have the parametric equations, we can use them to find the velocity and acceleration of the object. The velocity vector-valued function is represented by the first derivative of the position vector-valued function with respect to t. The acceleration vector-valued function is represented by the second derivative of the position vector-valued function with respect to t.

Let's solve the following problem step by step:

To solve this problem using parametric vector-valued functions, we first need to find the position vector-valued function:

r(t) = <x(t), y(t)> = <t^3 - 6t^2, 2t^2 - 4t>

Next, we need to find the velocity vector-valued function, which is the derivative of the position vector-valued function with respect to t:

v(t) = dr/dt = <dx/dt, dy/dt> = <3t^2 - 12t, 4t - 4>

To find the position at time t = 2 seconds, we substitute t = 2 into the position vector-valued function:

r(2) = <2^3 - 6(2)^2, 2(2)^2 - 4(2)> = <-4, 4>

This means the particle is at x=-4 units and y=4 units at t=2 seconds. Similarly, we can find the velocity of the particle at any given time by substituting the time into the velocity vector-valued function. For example, the velocity of the particle at time t = 2 seconds is:

v(2) = <3(2)^2 - 12(2), 4(2) - 4> = <4, 0>

This means that the particle has a velocity of 4 units/s in the x-direction and 0 units/s in the y-direction at time t = 2 seconds.

A **polar plane** is a two-dimensional coordinate system in which the position of a point is determined by the **distance from the origin (r)** and the **angle (theta ΞΈ**) between the positive x-axis and the line connecting the point to the origin, counterclockwise. The polar coordinates of a point **(r, ΞΈ)** in the polar plane are represented by an ordered pair of real numbers, where r is the distance from the origin and theta is the angle measured in radians.

A polar function is a function of the form:

r = f(ΞΈ)

where r is the distance from the origin to a point on the polar plane, and theta is the angle between the positive x-axis and the line connecting the point to the origin.

To find the **derivative** of a polar function, we can use the **chain rule **to derive a formula. It is helpful to memorize the formula, but you can also derive it during the test.

You can also **convert** between a polar function and a Cartesian function. To go from polar to Cartesian, use the first two formulas, and to go from Cartesian to polar, use the third formula.

The **area** of the region enclosed by a polar curve is given by the definite integral:

Where a and b are the limits of integration and r is the polar function. This integral is calculated by taking the product of 1/2 and the square of the polar function, and then integrating this expression with respect to theta from a to b.

It's important to note that this method of finding the area under a polar curve is only valid for **closed curves**, meaning the curve starts and ends at the same point. If it's not a closed curve, we have to find the area enclosed by the curve and a line connecting the start and the end of the curve.

The area of the region enclosed by **two polar curves** is given by the definite integral:

A = (1/2) β«(a,b) (R^2 - r^2) dΞΈ

Where a and b are the limits of integration, R is the equation of the outer curve and r is the equation of the inner curve. This integral is calculated by taking the difference of the square of the outer curve and the square of the inner curve and then integrating this expression with respect to theta from a to b.

This is similar to finding the integral between two curves in the Cartesian plane. Where you subtracted the bottom curve from the top curve, you'll now subtract the **inner curve** from the **outer curve.**

Browse Study Guides By Unit

πUnit 1 β Limits & Continuity

π€Unit 2 β Fundamentals of Differentiation

π€π½Unit 3 β Composite, Implicit, & Inverse Functions

πUnit 4 β Contextual Applications of Differentiation

β¨Unit 5 β Analytical Applications of Differentiation

π₯Unit 6 β Integration & Accumulation of Change

πUnit 7 β Differential Equations

πΆUnit 8 β Applications of Integration

π¦Unit 9 β Parametric Equations, Polar Coordinates, & Vector-Valued Functions (BC Only)

- 9.1 Defining and Differentiating Parametric Equations
- 9.2 Second Derivatives of Parametric Equations
- 9.3 Finding Arc Lengths of Curves Given by Parametric Equations
- 9.4 Defining and Differentiating Vector-Valued Functions
- 9.5 Integrating Vector-Valued Functions
- 9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions
- 9.7 Defining Polar Coordinates and Differentiating in Polar Form
- 9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve
- 9.9 Finding the Area of the Region Bounded by Two Polar Curves

βΎUnit 10 β Infinite Sequences & Series (BC Only)

π§Multiple Choice Questions (MCQ)

βοΈFree Response Questions (FRQ)

πBig Reviews: Finals & Exam Prep

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