[br]You will need strong command of mathematics to succeed in physics. These problems are meant to cause you to use that mathematics in ways that are similar to those required by physics. Hint: If you have never been a fan of word problems, please learn (or decide) to like them. After all, that's all that nature has for us! [br][br]The math needed to solve the first set of these problems comes from prior courses and the second set from this chapter. The problems are meant to cause you to think in a way required by physics. Draw pictures to aid visualization. A few of the problems will test your intuition. Have fun learning and extending yourself. [br][br]In this homework and every homework, you need to show all steps of the mathematics. The idea is to demonstrate command of the concepts and mathematics.[br][br]1. The earth rotates 360 degrees on its axis in one sidereal day. This means that the same SIDE (SIDEreal - not the real etymology) faces space after this period of time. This DOES NOT take 24 hours. What takes 24 hours is the time between one side of earth facing the sun and then rotating until the same side faces the sun again. This is called a solar day. Why do you suppose these are different? [br]2. Which should be longer, the sidereal day or the solar day, and why? [br]3. Please assume a circular orbit for the earth, and 365.24 solar days per year to calculate how long (to the nearest minute) a sidereal day should last. [br]4. Consider a rod placed vertically into the ground with exactly one meter above the ground sticking up. Assume you are someplace like Bonneville, UT (famous for auto racing) where the earth will be assumed perfectly level and smooth without hills or valleys or trees to worry about. How far must the stick be from you so that it becomes invisible from a vantage point of being on the ground (face or eyeball at ground level as if you're an ant)? Consider the curvature of the earth in this case. Earth's radius is [math]6.37\times10^6m.[/math] Do the calculation again for an eyeball height of 1.5m above the ground. [br]5. How far will the center of a ball travel if the ball rolls along a level surface without slipping and it rotates exactly once around its center.[br]6. While riding a bicycle, is the top of the wheel moving at the same speed, faster than or slower than the cyclist as measured by a pedestrian standing still.[br]7. How long is the shadow cast on level ground by a 3.0m long vertical stick if the sun is 30 degrees above the horizon? [br]8. What if the vertical stick is on a 5 degree slope with the sun directly uphill and still 30 degrees above the horizon on level ground?   [br]9. What mathematical function f(x) has a slope that is proportional to the value of x? [br]10. What mathematical function f(x) has a curvature proportional to the negative of the function itself? [br]11. What does taking the derivative of a function tell you about the function? Is a derivative a local property in the sense that you can define the derivative of a function f(x) at x? [br]12. What does integrating a function tell you? Is an integral a local property in the sense that you can define the integral of a function f(x) at x? [br]13. Integrate [math]f(x)=3x^2-2[/math] from x=0 to x=2. [br]14. Take the same function’s derivative and evaluate it at x=3.[br]15. If you first differentiate a function and then integrate it, are you going to get the same function back that you started with? [br][br]ANSWERS:[br]1) Because a solar day includes an over-rotation.[br]2) The solar day should be longer.[br]3) 23h 56m[br]4) 3569m (should be an arc segment); 7941m[br]5) A distance equal to the ball's circumference.[br]6) Faster... twice as fast.[br]7) 5.2m[br]8) 6.15m[br]9) a quadratic (parabola)[br]10) a sine or cosine function[br]11) its slope or rate of change. it is local.[br]12) the area under the curve. not local, but over a range.[br]13) 4[br]14) 18[br]15) no. there is always the loss of constant terms upon differentiation that are not recovered upon integration.[br]

Provide full solutions with all math steps shown.[br][br][list=1][*]Given the vector [math]\vec{a} = -3\hat{i} + 4\hat{j}[/math], find the magnitude and direction (angle with respect to the positive x-axis) of [math]\vec{a}[/math].[br][/*][*]A vector [math]\vec{b}[/math] has a magnitude of 10 and makes an angle of 30 degrees with the positive x-axis. Find the x and y components of [math]\vec{b}[/math].[br][/*][*]Find the unit vector in the direction of [math]\vec{v} = 2\hat{i} - \hat{j} + 3\hat{k}[/math].[br][/*][*]A particle's position vector is given by [math]\vec{r}(t) = (2t^3 - t)\hat{i} + (t^2 + 5)\hat{j} - (3t)\hat{k}[/math]. Find the particle's velocity vector [math]\vec{v}(t)[/math] and acceleration vector [math]\vec{a}(t)[/math] as functions of time.[br][/*][*]Given [math]\vec{c} = 5\hat{i} - 2\hat{j} + \hat{k}[/math], find the direction cosines for this vector.[br][/*][*]Integrate the following vector function with respect to time: [math]\int (t^2\hat{i} - 4t\hat{j} + 2\hat{k}) dt[/math]. Be clear about why the constant of integration needs to be a vector (show the math where this arises).[/*][*]If [math]\vec{a} = 2\hat{i} - \hat{j}[/math] and [math]\vec{b} = -3\hat{i} + 2\hat{j}[/math], find [math]2\vec{a} - 3\vec{b}[/math].[br][/*][*]A force vector is given by [math]\vec{F}(t) = (3\cos(t))\hat{i} - (2\sin(t))\hat{j}[/math]. Determine the change in this force vector between t=0 and t=[math]\frac{\pi}{2}[/math].[br][/*][*]A particle starts at rest at the origin. It has a constant acceleration of [math]\vec{a} = 2\hat{i} - \hat{j}[/math] (m/s[sup]2[/sup]). We'll soon see that to find its change in velocity vector during some time interval, we need to integrate acceleration over that interval. Please find the change in velocity between t=0 and t.[/*][*]Identify and describe what is wrong with the following equation and provide a corrected version (if applicable, or explain why no correction is possible): [math]\vec{v} = \frac{5}{\vec{a}}[/math] where [math]\vec{v}[/math] is velocity and [math]\vec{a}[/math] is acceleration.[br][/*][/list]Answers:[br][list=1][*][math]|\vec{a}|=5[/math], [math]\theta \approx 126.87^\circ[/math][br][/*][*][math]b_x = 5\sqrt{3} \approx 8.66[/math], [math]b_y = 5[/math][br][/*][*][math]\frac{2}{\sqrt{14}}\hat{i}-\frac{1}{\sqrt{14}}\hat{j}+\frac{3}{\sqrt{14}}\hat{k}[/math][br][/*][*][math]\vec{v}(t) = (6t^2 - 1)\hat{i} + 2t\hat{j} - 3\hat{k}[/math], [math]\vec{a}(t) = 12t\hat{i} + 2\hat{j}[/math][br][/*][*][math]\frac{5}{\sqrt{30}}, \frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}[/math][br][/*][*][math]\frac{1}{3}t^3\hat{i} - 2t^2\hat{j} + 2t\hat{k} + \vec{C}[/math] , where [math]\vec{C}[/math] is a constant vector of integration.[br][/*][*][math]13\hat{i}-8\hat{j}[/math][br][/*][*][math] -3\hat{i}-2\hat{j}[/math][br][/*][*][math]\vec{v}(t) = 2t\hat{i} - t\hat{j}[/math][br][/*][*]This is what's wrong: Dividing a number by a vector is undefined. This is because the result isn't scalar or vector, but some undefined object. This means that a correction is not possible.[br][/*][/list]