Please split this thread up into 3 separate ones if you want all 3 questions answered. One question per thread please.
edit: I didn't see that you answered them and wanted them checked. It's ok then. My bad.
So we came back from christmas break and jumped right into studying for midterms, so I don't remember how to do these. I did what I could, but can someone please help?
problems on the top, answers on the bottom
4) Water is draining at the rate of 48π ft^3/min from the vertex at the bottom of a conical tank whose diameter at its base is 40 ft and whose height is 60 ft.
a) find an expression for the volume of water in the tank in terms of its radius at the surface of the water.
b) at what rate is the radius of the water in the tank shrinking when the radius is 16 ft?
c) how fast is the height of the water in the tank dropping at the instant that the radius is 16 ft?
5) Let f be the function given by f(x) = 2x^4 - 4x^2 + 1
a) Find an equation of the line tangent to the graph at (-2, 17)
b) find the x and y coordinates of the relative maxima and relative minima. Verify your answer.
c) Find the x and y coordinates of the points of inflection. verify your answer.
6) Let f(x) = integral from 0 to x of [cos(t/2) + (3/2)]dt on the closed interval [0,4π]
a) Approximate F(2π) using four inscribed rectangles.
b) Find F'(2π)
c) Find the average value of F'(x) on the interval [0,4π]
c) no idea how to go about that
5) a) I know that you use limits.......
b) I know that the critical points are at -1, 0, and 1, but I don't know what to do from there
c) no idea how to go about that.
6) a) 6π
I'm fairly sure that I have 6 mostly wrong, as well as the other two, but at least I tried....can someone please help?
nope. first of all, your answer should be negative (why?) so there's no way this could be right.b) 768π
recall, you have
so now, continue
hint: use the expression you used to write h in terms of r. then perform the same procedure you did in part (b) on that expressionc) no idea how to go about that
recall that the tangent line is a straight line. it is of the form . here, as you know, is the slope, which is given by the value of the derivative at the point you want the tangent line to touch the curve
we can find the tangent line at a point by using the point slope form.
recall, the equation of a straight line through the point with slope can be written as
again, is your slope (evaluated at ), you are given
good job!b) I know that the critical points are at -1, 0, and 1, but I don't know what to do from there
now you have to test if they are max or min points or neither.
draw a number line, and plot the points -1, 0 and 1 on it in their relative positions. now you are going to test each region. pick a number less than -1, pick another between -1 and 0, another between 0 and 1, and another greater than 1. you are going to plug in these numbers into the first derivative and see if you get a positive or negative answer. positive means the function is increasing on that interval, negative means it's decreasing. now write a plus or minus over the respective intervals. you can also draw an arrow going up from left to right to mean positive (increasing) and an arrow going down from left to right to mean negative (decreasing)
it is probably obvious, but i will say it anyway
provided the function is continuous around the point, the point is:
(a) a maximum if the derivative is zero at the point, positive to the left of the point and negative to the right
(b) a minimum if the derivative is zero at the point, negative to the left of the point and positive to the right
(c) neither if neither (a) or (b) happens
note, you can also do the second derivative test here. you have to find the second derivative anyway to answer the next part, so you may prefer doing that. hopefully you remember the test.
points of inflection occur at points where the second derivative is zero, and it alternates in sign on either side of the point.c) no idea how to go about that.
example, if for a particular question you find the the second derivative is zero at x = 1, but to the immediate left of x = 1 it is positive and to the immediate right it is negative, then x = 1 is an inflection point. (of course, we are assuming the function is continuous in a neighborhood around x = 1)
My new answers
4)a)right 1st time
I don't think either of these are right....
b)min: (±1,-1)/max: (0,1)
6)a) I'm still getting 6π, can you please show me how to do this part?
c)right 1st time
as i said, your answers should be negative. secondly, are you sure about those 's
you thought wrong. these are correct.I don't think either of these are right....
b)min: (±1,-1)/max: (0,1)
i'd probably write (a) in the y = mx + b form though
note that you want to approximate6)a) I'm still getting 6π, can you please show me how to do this part?
now let , and recall from what you did with Riemann sums, the area is given by:
where the 's are the right endpoints. (you should do a sketch to realize you need the right end points)
so the approximation is:
and you know what is i suppose. it is the width of the rectangles. you should be able to figure it out easily, but there's a formula for it if you can't
...c)right 1st time