Chapter 6 Take Home Test
This take-home test is designed to assess your understanding of the concepts covered in Chapter 6. It is due at the start of class on Tuesday, February 13th. You are allowed to collaborate with your classmates, but remember that you must submit your own individual work, showing all steps clearly. Good luck!
Key Topics Covered
This test will cover the following key concepts from Chapter 6:
- Areas Between Curves (Section 6.1): Calculating the area between two or more curves by setting up and evaluating definite integrals. Remember to sketch the region first!
- Volumes of Solids of Revolution (Section 6.2 & 6.3): Using methods like the disk, washer, or shell method to find the volume of a solid generated by rotating a region around an axis. Pay close attention to the axis of rotation! Visualizing the solid will help.
- Work (Section 6.4): Computing the work done in various scenarios, such as lifting an object or pumping liquid out of a tank. Remember that work is force times distance ($W = F \cdot d$). Be careful with units!
- Applications to Interest (Section 6.5): Applying integration to model and calculate interest earned under variable rates. This includes understanding how to set up the integral to find the total interest earned over a specific period.
Sample Problems (Similar to Test Questions)
Areas Between Curves
Example: Find the area bounded by $y = 2x$, $y = 8$, and the y-axis.
Hint: First sketch the region. Then, determine the limits of integration and set up the appropriate definite integral.
Volumes of Solids of Revolution
Example: Consider the region bounded by $y = 4x$ and $y = x^2$. Set up an integral to compute the volume of the solid obtained by rotating this region about the x-axis.
Hint: Use the washer method. The volume element is given by $\pi (R^2 - r^2) dx$, where $R$ is the outer radius and $r$ is the inner radius.
Work
Example: A bucket that weighs 70 lbs when filled with water is lifted from the bottom of a well that is 60 feet deep. Compute the work required to lift the bucket to the top.
Hint: The work done is equal to the force (weight) times the distance. In this case, $W = 70 \cdot 60$.
Applications to Interest
Example: Suppose a variable interest rate is modeled by $I(t) = 0.06 + 0.01 \sin(\frac{2\pi t}{365})$ for $0 \le t \le 365$, where $t$ is measured in days. How much interest is earned after $t = 15.21$ days?
Hint: Integrate the interest rate function over the given time interval: $\int_0^{15.21} I(t) dt$.
Important Reminders
- Show All Your Work: Partial credit will be awarded based on the clarity and correctness of your solution steps.
- Check Your Answers: If possible, verify your results using different methods or by estimation.
- Manage Your Time: Don't wait until the last minute to start the test.
Good luck, and don't hesitate to ask if you have any clarifying questions before the due date!