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course MTH 279

Most students coming out of most calculus sequences won't do very well on these questions, and this is particularly so if it's been awhile since your last calculus-related course.So give it your best shot, but don't worry if you don't get everything.

I'm trying to identify areas on which you might need a refresher, as well as familiarize you with terminology and ideas that might not have been covered in your prerequisite courses.
Most of this is these questions are related to things you don't want to get distracted by when they pop up in your assignments.
Give me your best thinking, and I'll give you feedback, including a lot of additional explanation should you need it.
 

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Question: 

`q001. Find the first and second derivatives of the following functions:

• 3 sin(4 t + 2)

• 2 cos^2(3 t - 1)

• A sin(omega * t + phi)

• 3 e^(t^2 - 1)

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Your solution: 

•

First derivative: 4*3cos(4t+2) = 12cos(4t+2)

Second derivative: 4*12*(-1)*sin(4t+2) = -48sin(4t+2)

•

First derivative: 2*2*3*cos(3t-1)*[-sin(3t-1)] = -12cos(3t-1)sin(3t-1) = -12sin[2(3t-1)] = -12sin(6t-2)

Second derivative: -6*12sin(6t-2) = -72cos(6t-2)

•

First derivative: omega*Acos(omega*t + phi)

Second derivative: (-1) * omega * A* omega sin(omega * t + phi) = -A * omega^2 sin(omega * t + phi)

•

First derivative: 2t*3 e^(t^2 - 1) = 6t e^(t^2 - 1)

Second derivative: 2t * 6t e^(t^2 - 1) = 12t^2 e^(t^2 - 1)

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You need the product rule on this last one. Your other term will be (2t) ' * e^(t^2 - 1).

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confidence rating #$&*:8232; 

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Given Solution: 

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Self-critique (if necessary): I'm just really not too confident in my algebra...

 

 

@&

Your algebra looks pretty good so far.

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Self-critique rating: N/a

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Question:  
`q002. Sketch a graph of the function y = 3 sin(4 t + 2). Don't use a graphing calculator, use what you know about graphing. Make your best attempt, and describe both your thinking and your graph.

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Your solution: 

The graph has an amplitude of 3, a period of 2 pi / 4 = pi / 2, and is phase shifted 2 units to the left.  

 

 

confidence rating #$&*:8232; 

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Given Solution: 

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Self-critique (if necessary):ok

 

 

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Self-critique rating:n/a

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4t + 2 = 4 ( t + 1/2). The shift to the left is 1/2.

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Question: 

`q003. Describe, in terms of A, omega and theta_0, the characteristics of the graph of y = A cos(omega * t + theta_0) + k.

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Your solution: 

 A is again the amplitude, omega is the period factor where period = 2 pi / abs(omega), theta_0 is the phase shift (that shifts the graph either right or left), and k is the vertical shift that shifts the graph either up or down.

confidence rating #$&*:8232; 

@&

The phase shift is theta_0 / omega

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Given Solution: 

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Self-critique (if necessary): n/a

 

 

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Self-critique rating: n/a

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Question: 

`q004. Find the indefinite integral of each of the following:

• f(t) = e^(-3 t)

• x(t) = 2 sin( 4 pi t + pi/4)

• y(t) = 1 / (3 x + 2)

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Your solution: 

 

• f(t) = e^(-3 t)

Indefinite integral = -1/3 e^(-3 t) + c where c is a constant

• x(t) = 2 sin( 4 pi t + pi/4)

Indefinite integral = (-1/2)cos( 4 pi t + pi/4 ) + c

• y(t) = 1 / (3 x + 2)

 Indefinite integral = 1/3ln(3 x + 2 ) + c

 

 

confidence rating #$&*:8232; 

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Given Solution: 

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Self-critique (if necessary):n/a

 

 

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Self-critique rating:n/a

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Question: 

`q005. Find an antiderivative of each of the following, subject to the given conditions:

• f(t) = e^(-3 t), subject to the condition that when t = 0 the value of the antiderivative is 2.

• x(t) = 2 sin( 4 pi t + pi/4), subject to the condition that when t = 1/8 the value of the antiderivative is 2 pi.

• y(t) = 1 / (3 t + 2), subject to the condition that the limiting value of the antiderivative, as t approaches infinity, is -1.

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Your solution: 

Indefinite integral = -1/3 e^(-3 t) + c where c is a constant

so when t = 0, we have -1/3 + c = 2, so then c = 7/3

Indefinite integral = (-1/2)cos( 4 pi t + pi/4 )

so when t = 1/8, we have (-1/2)cos( 2 pi / 4 + pi / 4 ) = (-1/2)cos(3pi / 4 ) = -1/2 * 1/sqrt(2)… hm… I don't know where to go with this one to be honest.

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Your general antiderivative has an integration constant.

(-1/2)cos( 4 pi t + pi/4 ) + c

takes value 2 pi when t = 1/8, so

-1/ (2 sqrt(2) ) + c = 2 pi

and

c = 2 pi + 1 / (2 sqrt(2)).

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 Indefinite integral = 1/3ln(3 x + 2 ) 

I don't think my approach is correct here...

 

 

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Your indefinite integral is 1 / 3 ln(3 t + 2) + c.

As t -> infinity, so does ln(3 t + 2), so there is no finite value of c that would make this expression approach -1.

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confidence rating #$&*:8232; 

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Given Solution: 

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Self-critique (if necessary): n/a

 

 

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Self-critique rating: n/a

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Question: 

`q006. Use partial fractions to express (2 t + 4) / ( (t - 3) ( t + 1) ) in the form A / (t - 3) + B / (t + 1).

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Your solution: 

 Hm… I really don't remember partial fractions too well, but I'll take a crack at it.

 So first we set up an equation such that 2 t + 4 = B (t - 3 ) + A (t + 1), and we need to choose A and B so that the numerators are equal for every value t, in particular, the numerators must be equal for t = -1, 3

Then evaluate it when t = -1, 3,

for t = 3, we have 10 = B*0 + A(4), thus 4A = 10 and A = 10/4

for t = -1, we have 2 = B(-4) + A(0), thus B = -1/2

So the final form is 10/4 / (t - 3) + (-1/2) / (t + 1) = 10 / (4t - 12)) -1 / (2t + 2)

 

 

confidence rating #$&*:8232; 

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Given Solution: 

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Self-critique (if necessary):n/a

 

 

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Self-critique rating:n/a

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Question: 

`q007. The graph of a function f(x) contains the point (2, 5). So the value of f(2) is 5. 
At the point (2, 5) the slope of the tangent line to the graph is .5.

What is your best estimate, based on only this information, of the value of f(2.4)?

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Your solution: 

Newton's method I believe

y = f(x), in this case, y = f(x_0) + f'(x_0)(x-x_0) = f(2) + f'(2)(2.4-2.5) = 5 + 0.5*0.1 = 5 - 0.05 = 4.95

 

 

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At x = 2 the slope is .5.

To reach x = 2.4 you move .4 units in the x direction.

Moving .4 units in the x direction when the slope is .5, the y value changes by .5 * .4 = .2.

So your best estimate would be y = 5 + .2 = 5.2.

In the formula you quote, x_0 would be 2, x would be 2.4 and f '(2) is, as you say, .5.

Your x_0 = 2.5 is not correct; it's not sure where 2.5 came from.

You were on the right track, had the right formula. With careful reasoning, and maybe a picture, you could have concluded that y would change by .2 and reconciled this with the formula.

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confidence rating #$&*:8232; 

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Given Solution: 

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Self-critique (if necessary):n/a

 

 

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Self-critique rating:n/a

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Question: 

`q008. The graph of a function g(t) contains the points (3, 4), (3.2, 4.4) and (3.4, 4.5). What is your best estimate of the value of g ' (3), where the ' represents the derivative with respect to t?

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Your solution: 

 Really not sure how to do this one. I would probably find two slopes, one from (3, 4) to (3.2, 4.4) and one from (3.2, 4.4) to (3.4, 4.5), so that would be (4.4-4)/(3.2-3) = 0.4 / 0.2 = 2 and (4.5-4.4)/(3.4-3.2) = 0.1 / 0.2 = 1 / 2. So it would be somewhere between 1/2 and 2. I would average them, so (1/2 + 2)/2 = (1/2 + 4/2)/2 = 5/2*1/2 = 5/4

 

@&

That would be a very reasonable approach, but it doesn't account for the trend of the derivative.

Alternatively you could observe that the derivative is clearly decreasing, and conjecture that perhaps at x = 3, the derivative would be somewhat greater than its average value between x = 3 and x = 3.2.

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confidence rating #$&*:8232; 

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Given Solution: 

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Self-critique (if necessary): n/a

 

 

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Self-critique rating:n/a

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Self-critique rating:

#*&!

@&

Not bad overall. You appear to be prepared for this course.

Check my notes and let me know if you have questions.

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@&

Not bad overall. You appear to be prepared for this course.

Check my notes, inserted at various points in the document, and let me know if you have questions.

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