Answer Equivalence in Calculus Courseware

We know that oftentimes in calculus, there's more than one way to solve a problem. While some online systems don't allow for multiple correct answers, Hawkes Learning's courseware was built by subject matter experts who painstakingly went through examples to ensure students are given credit for equivalent answers. Marvin, one of our lead calculus content…

We know that oftentimes in calculus, there’s more than one way to solve a problem. While some online systems don’t allow for multiple correct answers, Hawkes Learning’s courseware was built by subject matter experts who painstakingly went through examples to ensure students are given credit for equivalent answers.

Marvin, one of our lead calculus content editors, explained why it’s so important to include equivalent answers in the courseware: “There are often different methods of solving, and we don’t want to penalize students for getting a correct answer. When that happens, students get frustrated and doubt themselves. We want to boost their confidence.”

Our calculus subject matter experts Marvin and Claudia shared a few examples that show our courseware giving credit for correct alternative answers.

Sample Problem from Trigonometric Integrals

The first two correct answers are generated using Method 1 of solving, while the next three are generated using Method 2 of solving.

Problem

Evaluate the indefinite integral ∫ 7tan(4x)sec6(4x)dx. Use C for the constant of integration. Write the exact answer. Do not round.

Correct Answer 1

Method 1: We can use u-substitution with u = sec(4x) after rewriting the integral as
7 ∫ sec5(4x) · sec(4x)tan(4x)dx. Note that the answer has the fraction 7/24 as the coefficient of the secant function.

clc3-1.png

Correct Answer 2

Method 1: We can use u-substitution with u = sec(4x) after rewriting the integral as
7 ∫ sec5(4x) · sec(4x) tan(4x)dx. Note that the answer has the secant function as part of the numerator of the answer.

clc3-2.png

Correct Answer 3

Method 2: We can use u-substitution with u = tan(4x) after rewriting the integral as
7 ∫ tan(4x) [1 + tan2(4x)]2sec2(4x)dx. Note that the answer has several terms with tangent and fractional coefficients.

clc3-3.png

Correct Answer 4

Method 2: We can use u-substitution with u = tan(4x) after rewriting the integral as
7 ∫ tan(4x) [1 + tan2(4x)]2sec2(4x)dx. Note that the answer has the fraction 7/8 factored out.

clc3-4.png

Correct Answer 5

Method 2: We can use u-substitution with u = tan(4x) after rewriting the integral as
7 ∫ tan(4x) [1 + tan2(4x)]2sec2(4x)dx. Note that the answer has the fraction 7tan2(4x)/8 factored out.

clc3-5.png

Correct Answers 6 & 7

If students rewrite the integrand in terms of sine and cosine and work it out correctly, credit is also given.

Below are two examples of a student answering the problem using cos(4x).

1clc.jpg

 

2clc.jpg


Sample Problem from The Chain Rule

This question shows the application of the Chain Rule, and the correct answer can be written in different ways as shown below.

Problem

Find the derivative of the function F(x) = – 3(13 + 2√x)-5.

Correct Answer 1

The student applies the Chain Rule and writes the last factor as 1/√x.

clc1-1.png

Correct Answer 2

The student applies the Chain Rule and writes the last factor as x -1/2.

clc1-2.png

Correct Answer 3

The student applies the Chain Rule and rewrites the square root of x in terms of fractional exponents.

clc1-3.png

Correct Answer 4

The student applies the Chain Rule and rewrites the whole answer as one fraction using the positive exponent 6 for the expression in parentheses.

clc1-4.png

Correct Answer 5

The student applies the Chain Rule and rewrites the answer as one fraction using the exponent of negative 6 for the expression in parentheses.

clc1-5.png


Sample Problem from Integration by Parts

Problem

Evaluate the integral ∫(t + 1)e4tdt. Use C for the constant of integration. Write the exact answer. Do not round. (Hint: Use an alternative method if integration by parts is not required.)

Correct Answer 1

The student applies integration by parts and writes the answer obtained by evaluating
uv – ∫ v du.

clc2-1.png

Correct Answer 2

The student applies integration by parts and writes the answer as one fraction with the common denominator and e4t factored out. clc2-2.png

Correct Answer 3

The student applies integration by parts and writes the answer with e4t factored out but no common denominator for the fractions. clc2-3.png


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Interested in seeing more of the calculus courseware? Contact us today at info@hawkeslearning.com or 1-800-426-9538 to get free access to the student courseware!