Solutions for study problems.

Hello, calculusians. Click on the images below to see solutions to the study problems. I hope that you have first tried them yourself.

If you have disagreements and/or questions, post them to the blog. You should answer each other's questions as I cannot guarantee how often I will check the blog. Feel free to stop by after your other finals if you are still struggling.











Don't forget to study related rates, implicit differentiation, and extrema. Practice some chain rule problems from your text or of your own creation.

Good Luck.

QOD:

Each one of these represents position, velocity and acceleration.

Acceleration is the derivative of velocity, and velocity is the derivative of position.

We will therefore note acceleration as the second derivative, velocity as the derivative and position as the function.

Key point on the graph:

We not that the slope of the green line approaches it's lowest quantity here, and that the y value of the red one reaches it's lowest value. Following that, the blue line reaches a y value of 0 when the red line's slope approaches zero.

Taking this point as representative of the function and its derivatives as a whole, then we assume that the green line is the function, the red line is the derivative, and the blue line is the second derivative.

They match up at all points. At the beginning, green (position)'s slope is zero, and red (velocity) is at zero, while it's slope is at it's lowest value -- and the green point is at it's lowest y value.

We then went over the function:

c(x) = 5000 + 10x + .05x^2

Where c is cost and x is designer glasses.

Asking for a) The average cost of designer glasses of 100 pairs, b) The marginal cost of 100 and c) The change in cost of 101 and 100.

a) asks for the plugging in of 100 as x and then dividing it by 100, finding the cost of producing each pair, at a production level of 100. Doing so:

(5000 + 10(100) + .05(100)^2)/100 = 65

It costs 65 to produce each pair of glasses at 100 glasses being produced.

b) asks for the change in cost at 100. We find this deriving. I used the power rule.

c(x) = 5000 + 10x + .05x^2

c'(x) = 5000(0)(x^-1) + 10(1)(x^0) + .05(2)(x^1)

c'(x) = 10 + .1x

We then substitute x for 100 to find the slope at that particular point.

c'(100) = 10 + .1(100)

c'(100) = 20

So the cost is increasing by 20 at that particular point.

c) asks for the difference between the cost at 101 and 100.

c(101) - c(100)

5000 + 10(101) + .05(101)^2 - 5000 - 10(100) - .05(100)^2

20.05

To our athletic compadre

Hey Max, whether you check this during your trip or when you get back I hope you guys did well and came back with your heads held high. I am going to try my hardest to post as nice as you do it but I am not sure if i am quite that blog savvy yet. Thank goodness there was no graphs! I really don't know how to write powers on this so I just put carrot ... sorry about that. Homework: Read- Pgs.128-129, Do questions #1, 3, 9, 11, 23, 25.

QOD

A bacteride was added to a growing population of bacteria. The size of the population in time t, in hours was given by: P(t) = (10^6)+(10^4)x-(10^3)x^2. This is 10 to the power of 6 plus 10 to the power of 4, x plus 10 to the power of 3, x squared. Find:

a)the rate of growth at 3 hours, 5 hours and 8 hours.

b)is the reate of growth increasing or decreasing at t = 7 hours?

a) In order to find the rate of growth we took the derivative of P(t) which is (10^4)-2(10^3)x which come out to: P'(t) = 10000-2000x We then plugged in each number... P'(3) = 10000-2000(3) = 4000 bacteria/hour , P'(5) = 10000-2000(5) = 0 bacteria/hour , P'(8) = 10000-2000(8) = -6000 bacteria/hour

b) In order to find out if the rate of growth was increasing or decreasing we had to (this might be a little confusing but you will understand it hopefully) find the rate of growth or the rate of growth. This would show if it is positive or negative, thus if it is increasing or decreasing. This is the same as d^2y / dx^2. This is finding the derivative of the derivative.

P''(t) = -2000 (beacause the derivative of P'(t) = 10000-2000x = -2000) P''(7) = -2000 (because there is no x so it is constant) bacteria/hour/hour Therefore it is decreasing (because it is negative) by 2000.

Part A tells us: If our starting point if 0, from 0 to 5 hours the population of bacteria is increasing. At t>5 the population of bacteria is decreasing. The maximum population is at t = 5 hours.

Part B tells us: As the population of bacteria decreases the rate of growth increases (in an absolute value sense).
_________________________________________________________________

What is the meaning of f '(a)?

• the slope of a tangent to f at x=a
• the limit of the slopes of the secants to f at x=a
• instantaneous rate of change of f at x=a
  

What is the meaning of f ''(a)?

• the derivative of f '(a)
• rate of change of the slope of f at x=a
• instantaneous rate of change of the instantaneous rate of change of f at x=a

-the value of f ''(a) tells you whether f '(a) is increasing, decreasing or constant

Higher Order Derivatives
1st derivative f '(x) or dy/dx
2nd derivative f ''(x) or (d^2)y/dx^2
3rd derivative f''''(x) or (d^3)y/dx^3
nth derivative f^n(x) or (d^n)y/dx^n

Pg. 133 #29
Cost of Production: a function that gives the cost associated with varying levels of production

Marginal Cost of Production:
"The change in cost when production in increased by 1 unit." - Definition from Mr. A

*Marginal cost of production always varies depending on where you are starting.
For example: If you are starting at zero prodution and want to change your unit to 1000 production, the change in cost is going to be very large because you need to pay for startup costs such as machinery. As opposed to if you are starting at 1000 to 2000 production the change in cost is not going to be as much because you already have payed the startup costs.

The equation for the average rate of change is your classic:

C2-C1 /X2-X1

Where c=cost, x=# of items produced. The numerator is the 2 costs compared (C1 being the starting production cost and C2 being your new production cost). The denomonator is the jump in production (X1 being your starting amount of items being produced and X2 being your new amount of items you wish to produce)

This is an average rate but in Calculus, we want instantaneous. How do you find this...... THE LIMIT!

The equation for the instantaneous rate of change is:

wow, that was super hard to write that and post it. haha. anyways...

This is the marginal cost when x=x1.

So what is marginal cost?

It is setting the level of production to maximize prodution.

Here is an example and some questions to try out:

Total cost in dollars of producing x designer eyeglass frames is given by:
c(x)=5000+10x=0.05x^2

*The cost of the function should be continually increasing.
*You should also see the fixed cost vs. variable cost. A fixed cost may be the cost of a machine to shape plastic, a variable cost may be say cost of screws for the frames because this depends on the amount of frames you are producing.

a)find the average cost of producing 100 frames
b)find the marginal cost at a production level of 100 frames
c)find the actual cost of increasing production from 100-101 frames
-actual cost is found by taking the cost of 100 and 101 and then you will know the cost of this increase.


Good Luck!

Scribe Post Evaluation Form

Here is a copy of the form that I will fill out and return to you after you make your scribe post. Use this post and the scribe post instructions (see sidebar) to guide you in writing your scribe post.

For Max/ Emily

QOD: A car starts at rest, rapidly accelerates at a constant rate for 2 sec, coasts for 2 sec, slowly decelerates at a constant rate for 4 sec, coasts 5 more sec, and then brakes to a stop with a constant acceleration.









d = accelerates with a curve for 2 sec
maintains slope for 2 secs
decreases with a curve for 4 secs
maintains slope for 5 secs
momuntarily maintains slope, but changes to a horizontal line

d' = accelerates at a constant rate of change, a line, for 2 secs
continues unto a horizontal line for 2 secs
decreases at a constant rate of change, another line, for 4 secs
continues unto a horizontal line for 5 secs
decreases into a horizontal line

d'' = horizontal line for 2 secs
hole, due to corner, turns to a horizontal line at y=0 for 2 secs
hole, due to corner, turns to a horizontal line at y=-.5 for 4 secs
hole, due to corner, turns to a horizontal line at y=0 for 5 secs
hole, due to corner, turns to a horizontal line at 7=-.25

Test Review Questions

Pg 173 #s 7, 55, 57, 58

Good Luck Everyone!!!

Hello Rachel and Connor

10/22
QOD

Suppose that f and g are both differenciable at x = 3 and that f(3) = 7, g(3)= 2, the derivative of (f)= -1 and the derivative of (g)=4. Using x=3 find:

a) d/dx (g x f)
b) d/dx (g/f)
c) d/dx (-2f-gf+3g)


*You should try and answer these and i will post the work below*
























Ok now that you have tried these i will now walk you throught them and give you the answers.

a) d/dx (gf) (gf)' = f'g + g'f

= -1(2) +4(7)

= -2+28

=26

b) d/dx (g/f) (g/f)'= g'f-gf'/f(squared)

= 4(7)-(-1)2/7(squared)

= 30/49

c) d/dx (-2f-gf+3g) (-2f-gf+3g)' = -2f'-(gf)'+3g'

= -2(-1)-26+3(4)

= -12

I am sure you guys got those pretty easy. On to todays lesson.


Today's Lesson


Find all domains where the function is indifferencaible.

The function below is a piecewise function.


absolute value of x where x is less than 2
x(squared) - 2 where x is greater than or equal to 2 and less than or equal to 5
f(x) x-7/x(squared)-49 where x is less than 5


What are the first basic things that we need to check?

1. Check problems that may occur in the three individual problems
2. One sided limit boundaries
3. One sided derivatives at boundaries


Work for the listed above

1. Absolute value of x has a cusp or a corner at x = 0 therefore we must consider that because the function takes on this identity everywhere less than 2 and because zero is less than 2 the function is indifferenciable at x = 0. Now we must look at x(squared) - 2. Because that function is a polynomial it is continuous everywhere and thus differentiable at every domain. 1/x+7 which is the simplified version of the third element of our function has a hole at x = 7, which makes it discontinuous and thus it also differentiable at x = 7.

2. Now lets check the one sided limit values at the boundaries.

l 2 l = 2(squared) - 2
2 = 2

5(squared)-2 does not equal 5-7/5(squared) -49

Therefore there is a corner at x=5

3. Ok now we must check one sided derivatives at boundaries. So...

d/dx ( lxl ) = d/dx (x) for x greater than 0 so d/dx (x) = 1 .

d/dx (x*squared*-2) = 2x

because we are observing the boundary x=2 approaching from the right, we must subsitute 2 in for x and that yields 4. This slope is not equal to the answer of one which is the slope at 2 when approaching from the left, thus we may concur that there is a corner there. The same may be done for the slopes at x = 5 but we already know it is indifferencialbe there.

Another piece of quiz review in todays lesson was the application of the power rool....i just gave you a hint for this one. Try it and the answer will be posted below! :]


3(x*squared*) / x*power of 3/4





























Answer: DAILY DOUBLE!!!.....




































JK GUYS! THE ANSWER IS... F' (X) = 15/4 *FOURTH ROOT OF (X)