✍️ Free Response Questions (FRQ)
Calculus Free Response Questions
👑 Unit 1: Limits & Continuity
1.5Determining Limits Using Algebraic Properties of Limits
1.6Determining Limits Using Algebraic Manipulation
1.10Exploring Types of Discontinuities
1.11Defining Continuity at a Point
1.12Confirming Continuity over an Interval
🤓 Unit 2: Differentiation: Definition & Fundamental Properties
2.4Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist
🤙🏽 Unit 3: Differentiation: Composite, Implicit & Inverse Functions
3.0Unit 3 Overview: Differentiation: Composite, Implicit, and Inverse Functions
3.1The Chain Rule
3.3Differentiating Inverse Functions
3.4Differentiating Inverse Trigonometric Functions
👀 Unit 4: Contextual Applications of the Differentiation
4.2Straight-Line Motion: Connecting Position, Velocity, and Acceleration
4.4Intro to Related Rates
4.6Approximating Values of a Function Using Local Linearity and Linearization
✨ Unit 5: Analytical Applications of Differentiation
5.0Unit 5 Overview: Analytical Applications of Differentiation
5.2Extreme Value Theorem, Global vs Local Extrema, and Critical Points
5.3Determining Intervals on Which a Function is Increasing or Decreasing
5.4Using the First Derivative Test to Determine Relative (Local) Extrema
5.7Using the Second Derivative Test to Determine Extrema
🔥 Unit 6: Integration and Accumulation of Change
6.11Integrating Using Integration by Parts
💎 Unit 7: Differential Equations
7.0Unit 7 Overview: Differential Equations
7.7Finding Particular Solutions Using Initial Conditions and Separation of Variables
🐶 Unit 8: Applications of Integration
8.1Finding the Average Value of a Function on an Interval
8.2Connecting Position, Velocity, and Acceleration of Functions Using Integrals
8.3Using Accumulation Functions and Definite Integrals in Applied Contexts
8.4Finding the Area Between Curves Expressed as Functions of x
8.5Finding the Area Between Curves Expressed as Functions of y
8.6Finding the Area Between Curves That Intersect at More Than Two Points
8.7Volumes with Cross Sections: Squares and Rectangles
8.8Volumes with Cross Sections: Triangles and Semicircles
8.9Volume with Disc Method: Revolving Around the x- or y-Axis
8.10Volume with Disc Method: Revolving Around Other Axes
8.11Volume with Washer Method: Revolving Around the x- or y-Axis
🦖 Unit 9: Parametric Equations, Polar Coordinates & Vector Valued Functions (BC Only)
9.0Unit 9 Overview: Parametric Equations, Polar Coordinates, and Vector-Valued Functions
9.1Defining and Differentiating Parametric Equations
♾ Unit 10: Infinite Sequences and Series (BC Only)
10.0Unit 10 Overview: Infinite Series and Sequences
10.1Defining Convergent and Divergent Infinite Series
10.6Comparison Tests for Convergence
10.7Alternating Series Test for Convergence
10.1110.11 Finding Taylor Polynomial Approximations of Functions
10.14Finding Taylor or Maclaurin Series for a Function
🧐 Multiple Choice Questions (MCQ)
⏱️ 2 min read
June 8, 2020
Now that you can find the area between curves, you should have no problem finding the volumes of 3-dimensional shapes. In order to find the area of a shape using a square cross section, take a look at the example below.
Let’s say we have a function x = 4 - y^2 and we know that the cross sections are squares. Because the height and width of squares are the same, finding the volume is quite easy. Before, we were finding the area by adding up a bunch of infinitely thin slices. Now, we are finding the volume by adding up a bunch of infinitely thin squares.
The area of one square cross section is ((4 - y^2) - -(4 - y^2))^2. This can be simplified to (2(4 - y^2))^2. This is a simple right minus left represented in topic 8.5. Now we integrate from the intersection points found by setting y = 0.
Like before with squares, we can use rectangles to find the volume of an object. The area of a rectangle is A = lw, where l is length and w is width. In each case, once we figure out l and w, we can integrate between our bounds to find the volume.
Problem Courtesy of KhanAcademy
We're given the function y = x^3/2 and told to find a volume with rectangular cross sections perpendicular to the y-axis. This is an important distinction. This means that the length of each rectangle is NOT x^3/2 (y), but rather (2y)^(1/3) [this is obtained by solving for x], since our rectangles are in terms of y, not x. Now, we just need to find l. Luckily, we're simply given this in the problem. Therefore, our integral will be the integral from 0 to 4 of (2y)^(1/3) * y/2.
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