Trigonometric Functions and Area Calculations

Trigonometric Functions and Area Calculations

Assessment

Interactive Video

Mathematics

10th - 12th Grade

Hard

Created by

Emma Peterson

FREE Resource

The video tutorial explains how to find the area of a region bounded by two polar curves: r = 5 sin(θ) and r = 4. It covers the use of integration to calculate the area between these curves, considering symmetry to simplify the process. The tutorial details setting up the integral, finding intersection angles, and performing calculations to arrive at the final area. The process involves using trigonometric identities and a calculator for approximations, ultimately finding the area to be approximately 3.7477 square units.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the main objective of the problem discussed in the video?

To find the area inside the red circle.

To find the area inside the blue circle and outside the red circle.

To find the intersection points of the two curves.

To find the perimeter of the blue circle.

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which formula is used to find the area between two polar curves?

Area = 2πr

Area = ∫(r₂² - r₁²) dθ

Area = ∫(r²) dθ

Area = πr²

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why can the integration be stopped at π/2 radians in this problem?

Because the area is symmetrical across the x-axis.

Because the area is symmetrical across the y-axis.

Because the curves do not intersect beyond π/2.

Because π/2 is the maximum angle for polar coordinates.

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is the angle α determined in the context of this problem?

By measuring directly from the graph.

By setting the two equations equal and solving for θ.

By finding where r = 0.

By using the cosine function.

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the value of θ when solving for α using inverse sine?

θ = 1 radian

θ = π/4

θ = inverse sine of 4/5

θ = π/2

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What substitution is used to simplify the integral of sine squared θ?

sine squared θ = 2 sine θ cosine θ

sine squared θ = cosine squared θ

sine squared θ = 1/2(1 - cosine 2θ)

sine squared θ = 1 - cosine θ

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the antiderivative of cosine 2θ used in the solution?

1/2 cosine θ

1/2 sine 2θ

cosine 2θ

2 sine θ

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