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

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

These are all classic L'Hopital's Rule questions.  Study my Lesson 11 thoroughly to prepare.

Study Lesson 14 but only study pages 157-161 (questions 1 to 3), pages 164-166 (question 5, parts (a), (b) and (c), and pages 168-175 (question 6) at this time.  You are not ready to do the integral applications of Parametric Equations yet.

They want you to change this into an ordinary "y as a function of x" kind of equation.  For example, if I had the parametric equations of x=2t and y=t^2+1, I could use the first equation to solve for t:

Since x=2t, it follows that t=x/2.  Then I could sub that into the second equation in place of t.  So, I would get y=(x/2)^2+1 or y=x^2/4 + 1, and I have now written the parametric equations in Cartesian form.

The thing you want to notice here, is how similar the parametric equations for x and y are. 

Hint: Square both sides in that x equation, and you get something very similar to the y equation.  You can then do a substitution to get an equation in the form "y= a function of x."

The hint they give you is related to the original parametric equations you were given.  We must notice that x could never be negative because x is a square root function.  For example, when t=0, x= sqrt(3), a positive number.  When t=1, x=sqrt(4)=2, a positive number.  The square root of any nonzero number is always positive! You would never get a negative answer for x.  That's what they mean by the range of x.

The wording may be confusing you, because we usually talk about the domain of x and the range of y.  But, in parametric equations, we state x as a function of t, and y as a function of t.  Our independent variable is t.  Both the x and y coordinates are dependent variables because their values depend on t.  We get the domain of independent variables and the range of dependent variables.  So, in this context, we discuss the range of x and the range of y.

Of course, we must also be mindful of the usual restrictions on domain or range.  Can't have values that cause a bottom zero (relevant here) or negative values inside a square root (irrelevant here, because there won't be a square root in your answer to part (a) if you did it right).

This is probably most easily done by just drawing the graph you figured out in part (a), keeping in mind the restrictions on x values noted in part (b).  Plot about three points, and make sure you solve the limit as x approaches 0+ to understand what the graph is doing there.

This draws the graph but does not tell you the direction arrows.  Use the parametric equations to get the correct direction.  Sub in about three values of t to get that straight.  The points you get should tally with the graph you have already drawn, but now you will clearly see the direction the points are traveling.

Similar to my Lesson 14, #1.  There is some factoring and canceling available that you must do to simplify the answer as they have requested.

Similar to my Lesson 14, #2 and #3.

Study Lesson 15 but only study pages 176-185 (question 1) and page 188 (question 4) at this time.  You are not ready to do the integral applications of Polar Curves yet.

Plot the two given points on a graph, keeping in mind that those are polar coordinates (r, θ), so you rotate to the designated angle line, then travel the prescribed distance along that angle line from the origin to plot the point.

Once you have plotted those two points, you can draw a triangle using the two given points and the origin as the three vertices.  The distance between the two points is a side of that triangle.  But, this is not a right triangle! The most efficient way to solve this distance is the law of cosines.

Again, make sure you have studied Lesson 15, #1 before attempting this question.

Sketch the polar graphs 1=r, rather than 1<=r and rsinθ=1, rather than rsinθ<1 for the given domain of the angle θ.  As always with polar curves, just make a table of values and plot the points.  Once you have sketched the curves, you will shade the appropriate regions that are indicated by the inequalities.  Shade the region that satisfies all the conditions.  For example, 1<=r is saying that the radial distance r must be greater than or equal to 1 (since 1 is less than or equal to r), and since r=1 is just a circle of radius 1, r>=1 is all the points outside of that circle's circumference.

Note also that π/4 <= θ < π/2 is not only telling you what angles to put on your table of values, but that is also giving you lines to draw on your sketch.  You should draw a diagonal line at the angle of π/4 and another line at the angle of π/2 (which is simply the postive y axis).  Technically that second line should be indicated by a dotted line since it is up to but not including π/2.  Any shading you do for the regions of the two other graphs you were given must be restricted to that triangular "pizza slice" between the angles of π/4 and π/2 and radiating out from the origin.

Again, make sure you have studied Lesson 15, #4 before attempting this question.

The key property of integrals they are referring to here is the concept I discuss in Lesson 2, page 41.  A definite integral can be interpreted as an area.  Specifically, the definite integral from a to b of f(x) dx, is computing the area between y=f(x) and the x-axis bounded by the vertical lines x=a and x=b.

Note that both definite integrals are very similar.  Same endpoints (x=0 and x=1), and almost the same function, just a different power.  Graph y=(1+x^2)^(1/3) and y=(1+x^2)^5 between x=0 and x=1.  This is easily done by just plotting three points to get a rough sketch, x=0, x=1, and another x value between 0 and 1.  I would draw both curves on the same graph.  Then, we can easily see which would have the greater area, and so prove the inequality.

Same idea, except this time you only have one definite integral, so only one graph to draw.  Can you see how that graph obviously has an area greater than or equal to 4π/3?