Chp 10 Review worksheet:
4c) To find this volume, you need to find the equation for the area of one cross section, then integrate it from x = 0 to x = 3. Each cross section will be shaped like a "washer", because there will be a "big radius" out to the y = 3/2x^2 function, and a "small radius" out to the x-axis. The portion from y = -3 up to the x-axis will be a hole, because the rotated area will not cover this part.
Everything will be in terms of x, because the area is rotating around a horizontal axis (like the x-axis, but lower on the graph).
The integral's bounds will be 0 and 3.
The area of the "big circle" will be pi(3/2x^2 + 3)^2. The radius is formed by 3/2x^2, which is the distance from the x-axis to the function, and by adding +3, which is the distance from the x-axis to the center of rotation.
The area of the "small circle" will be pi(3)^2. The radius is formed by 3, which is the distance from the x-axis to the center of rotation.
(integral sign from 0 to 3) [ pi(3/2x^2 + 3)^2 - pi(3)^2 ] dx
Answer: 598.002
5) Keep in mind the big ideas of what you're trying to do here: You've been given acceleration data, and you need to find velocity. (This means you need to integrate!). Then, you'll have velocity data, and you need to find displacement. (This means you need to integrate again!)
Because we just have data, and not an equation, we can't find the anti-derivative of acceleration. Instead, we can integrate by thinking about this like a Riemann Sum.
Imagine plotting all the a(t) points on a graph. If we needed the integral (velocity), we could draw midpoint Riemann Sum rectangles to come up with a decent estimate.
For the first interval, we'd have a rectangle from t = 0 to t = 2 (width of 2), whose height can be found by averaging the data points on either side of it (height of 6.5). This first Riemann Sum rectangle will have an area of 2*6.5 = 13.
So, we just integrated acceleration by making a Riemann Sum rectangle to find the change in velocity. We found that change in velocity to be 13 ft/sec, which means that the total velocity at t = 2 sec is now 23 ft/sec (add the 13 to the starting velocity of 10).
Continue in this fashion to find the next velocities. Here are the calculations:
Change in velocity at 4 sec:
2*5 = 10 ft/sec
...this makes the actual velocity at this moment = 33 ft/sec.
Change in velocity at 6 sec:
2*7 = 14 ft/sec
...this makes the actual velocity at this moment = 47 ft/sec.
NOW, do the same thing, but integrate the velocity numbers in order to find displacement. We'll imagine having a graph of the v(t) data points, and creating Riemann Sum rectangles with these to find the change in displacement.
Change in displacement at 2 sec (find the 16.5 number by averaging the velocity at 0 and 2):
2*16.5 = 33 ft
...this makes the actual displacement at this moment = 33 ft.
Change in displacement at 4 sec:
2*28 = 56 ft
...this makes the actual displacement at this moment = 89 ft.
Change in displacement at 6 sec:
2*40 = 80 ft
...this makes the actual displacement at this moment = 169 ft.
Chp 10 Review book work (pg. 548):
R4) This problem forms a right triangle, where the rate at which Rover is moving is equal to the rate at which the hypotenuse is increasing (because Rover is pulling the tablecloth "hypotenuse"). This will be the same rate the glass is moving, because everything attached to the tablecloth will be moving at the same rate.
Start by setting up the Pythagorean Theorem:
a^2 + 70^2 = h^2
...where "a" is the horizontal distance from the table to Rover, "70" is the height from Rover's mouth to the top of the table, and "h" is the hypotenuse (tablecloth diagonal distance).
Find the implicit derivative:
2a * da/dt = 2h * dh/dt
a * da/dt = h * dh/dt
Here's the information we know:
da/dt = 20 cm/sec (because Rover is moving horizontally along the "a" side of the triangle)
h = 200 cm (because the glass will fall off when the tablecloth is fully pulled off of the table)
We want to know:
dh/dt
We'll also need to plug in a value for "a" in the equation, but we can find this by plugging "h" back into the original Pythagorean Theorem:
a^2 + 70^2 = 200^2
a = 187.350 cm
Plug the values you know into your derivative equation, and solve for dh/dt:
187.350 * 20 = 200 * dh/dt
dh/dt = 18.735 cm/sec
To answer the final part of this question:
"The instant the glass reached the table's edge, it was traveling 18.735 cm/sec, which is 1.265 cm/sec slower than Rover is moving."
Sec 10.4 (pg. 519) help:
7) This problem wants you to analyze how the length, width, and area of a rectangle is changing. Let's start with the equation for area:
A = L * W
Then, let's list the information we already know, and the information we want to know:
We know...
dL/dt = 3 ft/min
dW/dt = -2 ft/min
L = 50 ft
W = 20 ft
We want to know...
dA/dt ???
So, they gave us enough information that we don't need to worry about using some other equation to substitute and get rid of some of our current variables. We can just do an implicit derivative with the simple A = L*W equation. Keep in mind - use PRODUCT RULE!
Derivative of area equation with respect to time t:
dA/dt = dL/dt * W + L * dW/dt
Now, plug in the values they gave us:
dA/dt = 3 * 20 + 50 * -2
dA/dt = -40
To figure out the units, realize that A is ft^2 and t is min, so dA/dt is ft^2/min. Also, if dA/dt is negative, that means that our area is decreasing.
Answer: The area of the rectangle is decreasing at 40 ft^2/min
13a) In this problem, keep in mind that as the top of the ladder slides down, the attached “weight” slides up at the same rate. So, we can just look at the height of the ladder on the wall, because all the matching information for the weight will have the same values, but negative. (i.e. if the ladder is sliding downward at 2 ft/s, then the weight is sliding upward at 2 ft/s).
Let x = distance from bottom of ladder to wall (the horizontal side of the triangle).
Let y = distance from top of ladder to floor (the vertical side of the triangle).
This lets you set up the Pythagorean Theorem as a starting equation:
x^2 + y^2 = 20^2
x^2 + y^2 = 400
Find the implicit derivative with respect to time t:
2x * dx/dt + 2y * dy/dt = 0
Divide the entire equation by 2, and subtract the term with x to the other side:
y * dy/dt = -x * dx/dt
Divide both sides by y:
dy/dt = -x/y * dx/dt
Then, it wants us to write the equation “as a function of the distance the bottom of the ladder is from the wall”, or in other words, “in terms of x”. We need to solve our original equation for y, then substitute this in for y in our derivative equation:
x^2 + y^2 = 400
y^2 = 400 - x^2
y = sqrt(400 - x^2)
Then substitute into dy/dt:
dy/dt = -x/sqrt(400 - x^2) * dx/dt
BUT, we know the rate at which the weight moves is the negative value of the rate at which “y” (the top of the ladder) moves. So, your final equation for the velocity of the counterweight will be the negative version of the equation we just found. (In simple terms…drop the negative sign to make your equation positive).
dy/dt = x/sqrt(400 - x^2) * dx/dt
15) This one was hard to explain by typing it out, so I made a video for you:
Educreations video: Sec 10.4 Problem #15a
Sec 10.2 (pg. 509) help:
13) Here's the general idea:
If you have acceleration data, and want to get velocity data, you need to take the integral of the acceleration data by using Riemann Sums. It's always more helpful to do these problems after graphing the points!
Try drawing at least the first couple points of the acceleration data: (0, 1.2) (5, 4.7) (10, 2.9). Then draw the mid-Riemann Sum rectangle. The first rectangle should have width 5, and height 2.95 (which is the average of 1.2 and 4.7). Find the area of this rectangle: 5 * 2.95 = 14.75.
This is the integral of the acceleration data from 0 sec to 5 sec, so this is what the velocity is at 5 sec: 14.75 mph
Next rectangle is the same type of math:
Width * Mid-Height = Velocity increase in that interval
5 * 3.8 = 19 mph increase from 5 sec to 10 sec
To actually find the velocity at 10 sec, we'd add these two answers together.
Velocity increase from 0 to 5: 14.75 mph
Velocity increase from 5 to 10: 19 mph
Total velocity at 10 sec: 33.75 mph
There's more to do after this obviously, but this is a start.
