Calculus I, Fall 2012 (http://www.math.nthu.edu.tw/˜ wangwc/)

Homework Assignment for Chap 06

1. Section 6.1: Problems 45, 47, 59, 65.

2. Section 6.2: Problems 28(c), 32, 45, 46.

3. Section 6.3: Problems 41, 50, 51, 57, 65, 77, 87, 89, 91, and as many as time permits in problems 31-40, 85-90.

4. Section 6.4: Problems: 9, 19, 25, 41, 47, 57, 71.

5. Evaluate lim_{n→∞} ∑_{k= \frac{n}{2}}^{n} \frac{1}{k}. Hint: try to express it in terms of \frac{k}{n}.

6. Section 6.6: Problems: 17, 21, 27, 33, 37-46, 51, 54, 55.

7. Section 6.7: Problems: 5, 7, 8, 11, 12, 17(a), 18.

8. Section 6.9: Problems: 9, 13, 21, 23, 31, 33, 39, 43, 57, 61, 67, 71, 79.

9. Show that

\frac{d csc^{−1} y}{dy} = \frac{−1}{|y|\sqrt{y^{2} − 1}}, |y| > 1

Explain why the negative sign is chosen. You will need to start from the ‘restricted’ domain of csc to see this.

10. Section 6.10: Problems: 15, 31, 43, 67.

For Section 6.10, skip the inverse hyperbolic functions part and concentrate on hyperbolic functions and calculus related to them.

11. Section 6.11: Problems: 17, 21, 23, 27.

12. Verify that both sinh x are cosh x are solutions of y′′ = y and then solve for y′′ = y, y(0) = 1, y′(0) = 2.

13. Verify that both e^{2x} and xe^{2x} are solutions of y′′ − 4y′ + 4y = 0, therefore so is the combination a_{1}e^{2x} + a_{2}xe^{2x}. This is an example of the multiple root case: (λ − 2)^{2} = 0. You can either verify by direct differentiation, or try to look for solutions of the form z(x)e^{2x} and find that this leads to z′′ = (y(x)e^{−2x})′′ = 0.

14. Verify by direct differentiation that

\frac{d}{dx}(cos(kx) + i sin(kx)) = ik(cos(kx) + i sin(kx))

This is a good explanation why one defines exp(ikx) to be cos(kx) + i sin(kx).