Don't have my book or audio?  You can download a free sample of my book and audio lectures containing Lessons 1 and  2:

Did you read my tips on how to study and learn Math 1500?  If not, here is a link to those important suggestions:

Did you see my tips for Assignment 1? Click here.

Did you see my tips for Assignment 2? Click here.

Did you see my tips for Assignment 3? Click here.

Did you see my tips for Assignment 4? Click here.

These are tips for the assignments in the Distance/Online Math 1500 course, but I strongly recommend that you do this assignment as homework even if you are taking the classroom lecture section of the course.  These assignments are very good (and challenging) practice. 

Here is a link to the actual assignment, in case you don't have it:

Study Lesson 10 (Max/Min Word Problems)  and Lesson 11 (Antiderivatives and Integrals) from my Intro Calculus book to prepare for this assignment.

First, make sure you have read my previous email about finding the domain in Max/Min Word Problems.  Be sure you do state the domain as part of your answer to the first two questions:

This question is quite unusual.  Be sure to sketch a graph of the two given curves to help you visualize the problem.  Remember that the problem is restricted to x in [-3,1].  That is your domain by the way!  

The sketch is an easy matter of just plotting the two given endpoints for the line, and maybe one or two extra points between to help you draw the parabola.

Since you want the maximum vertical distance, that is just D = y1 - y2, where y1 is the higher of the two curves in the region you sketched.  Substitute the two given equations in place of y1 and y2, and you now have your equation for D (that will be what I refer to as the "Q equation."

Similar to my question 6 in Lesson 10.  But, your base is a rectangle rather than a square.  What would that mean about your labels for the dimensions?

This is just standard antiderivative and indefinite integral stuff like what I am doing in my Lecture Problem 1 in Lesson 11. 

This is just standard antiderivative and integral stuff applied to velocity and acceleration, like what I am doing in my Lecture Problem 9 in Lesson 11. 

This question is NOT to be solved using antiderivative formulas! You couldn't solve it that way anyway. 

First, split the problem up into two separate definite integrals using properties of antiderivatives.  You can say it is the definite integral from -4 to 4 of x minus the definite integral from -4 to 4 of the square root expression.  Don't forget to put dx at the end of each integral!

For the integral of x, you have to draw a graph of f(x) = x from x=-4 to x=4, then find the area between the vertical lines x= -4 and x=4, between the graph of f(x) and the x-axis using geometric methods.  You should notice the shape is a couple of triangles, so you can find the area by using the area of a triangle. Remember! Area below the x-axis is negative.  In fact, if you think about it, the answer to this definite integral is pretty obvious.

For the other integral, draw the graph of f(x) = square root of (16- x^2) from x=-4 to x=4 and find the area between the curve and the x-axis.  Hint: the graph is a semicircle.

Once again, you can solve the integrals in part (a) by computing areas.  For example, g(3) is the integral from 0 to 3 according to their definition of g(x).

The best way to understand parts (b) and (c) is to do part (d) first.  Plot your answers from part (a) on a graph to sketch g.  For example, g(3) tells you the y value when x=3 for the graph of g.