1. Problem (a): Find $$\lim_{h \to 0} \frac{\sin\left(\frac{\pi}{2} + h\right) - 1}{h}$$. 2. Use the identity $$\sin\left(\frac{\pi}{2} + h\right) = \cos h$$. 3. Rewrite the limit as $$\lim_{h \to 0} \frac{\cos h - 1}{h}$$. 4. Recall that $$\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$$ because $$\cos h \approx 1 - \frac{h^2}{2}$$ near 0, so numerator is approximately $$-\frac{h^2}{2}$$ which goes to 0 faster than $$h$$. 5. Therefore, the limit for (a) is $$0$$. 6. Problem (b): Find $$\lim_{h \to 0} \frac{\csc(x + h) - \csc x}{h}$$. 7. Recognize this as the definition of the derivative of $$\csc x$$ at $$x$$. 8. The derivative of $$\csc x$$ is $$-\csc x \cot x$$. 9. Therefore, the limit equals $$-\csc x \cot x$$. 10. Problem 31: Ladder problem. 11. Given ladder length $$10$$ ft (hypotenuse), angle $$\theta$$ with horizontal, height $$x$$ is opposite side. 12. Relation: $$x = 10 \sin \theta$$. 13. Differentiate with respect to $$\theta$$: $$\frac{dx}{d\theta} = 10 \cos \theta$$. 14. Evaluate at $$\theta = 60^\circ$$: $$\frac{dx}{d\theta} = 10 \cos 60^\circ = 10 \times \frac{1}{2} = 5$$ ft/degree. 15. Problem 32: Airplane distance rate. 16. Height $$= 3800$$ ft, angle $$\theta$$ at point P, distance $$s$$ is hypotenuse. 17. Relation: $$\sin \theta = \frac{3800}{s} \Rightarrow s = \frac{3800}{\sin \theta}$$. 18. Differentiate with respect to $$\theta$$: $$\frac{ds}{d\theta} = 3800 \cdot \frac{-\cos \theta}{\sin^2 \theta} = -3800 \frac{\cos \theta}{\sin^2 \theta}$$. 19. Evaluate at $$\theta = 30^\circ$$: $$\cos 30^\circ = \frac{\sqrt{3}}{2}, \sin 30^\circ = \frac{1}{2}$$. 20. Substitute: $$\frac{ds}{d\theta} = -3800 \times \frac{\frac{\sqrt{3}}{2}}{\left(\frac{1}{2}\right)^2} = -3800 \times \frac{\frac{\sqrt{3}}{2}}{\frac{1}{4}} = -3800 \times 2 \sqrt{3} = -7600 \sqrt{3}$$ ft/degree. 21. The negative sign indicates $$s$$ decreases as $$\theta$$ increases. 22. Problem 33: Searchlight problem. 23. Distance from searchlight to building $$= 50$$ m, angle $$\theta$$, vertical height $$D$$. 24. Relation: $$\tan \theta = \frac{D}{50} \Rightarrow D = 50 \tan \theta$$. 25. Differentiate with respect to $$\theta$$: $$\frac{dD}{d\theta} = 50 \sec^2 \theta$$. 26. Evaluate at $$\theta = 45^\circ$$: $$\sec 45^\circ = \frac{1}{\cos 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$$. 27. Substitute: $$\frac{dD}{d\theta} = 50 \times (\sqrt{2})^2 = 50 \times 2 = 100$$ m/degree. Final answers: (a) $$0$$ (b) $$-\csc x \cot x$$ 31. $$5$$ ft/degree 32. $$-7600 \sqrt{3}$$ ft/degree 33. $$100$$ m/degree