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Implicit Differentiation

Q:  Using implicit differentiation, find dy/dx of

y/(x+6y)=x7-9

at the point (1,-8/49)

A:

Taking the derivative of the right side of the equation is fairly basic, taking the derivative of the left side, on the other hand, is harder.  The left side involves a quotient rule.  I am going to ignore the right side of the equation for now and just deal with the left:

y/(x+6y)

The quotient rule tells you (in some form / notation or another):

(bottom)*(deriv. of top) – (deriv of bottom)*(top) all divided by (bottom) squared.  [You may have learned this with f’s and g’s and f ‘ and g ‘ — it all says the same thing]

So:

Top = y

Bottom = x + 6y

Derivative of Top = dy/dx = y’ [whatever notation you are using]

Derivative of Bottom = 1 + 6(dy/dx) = 1 + 6y’

Continue reading Implicit Differentiation

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Use the area of a cirle to find the diameter

Q:  The area of a circle is 16π cm squared.  What is the diameter of the circle?

Answer:

First we need to know the area formula for a circle… This is:

A = π*r2  [r is the radius]

Since the area is given to us, we can plug that in to the formula to get:

16*π = π*2

The only variable is r, so we can use algebra to solve for r like so:

Divide both sides by π (which makes the π’s cancel out):

16*π  = π*r2

16 = r2

Now, take the square root of both sides to solve for r:

Continue reading Use the area of a cirle to find the diameter

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Find solutions to a Trig Function

Q:  Find all solutions for X where 0 ≤ X < 360 degrees and 9*cos(X) – 5 = 0.

Answer:

Okay, the 0 ≤ X < 360 just tells you we are looking for answers in one complete circle (no more, no less).

So, we first need to solve for X like so:

9*cos(X) – 5 = 0

9*cos(X) = 5

cos(X) = 5/9

X = cos-1(5/9)

Plug this into your calculator and you get:

X = 56.25 degrees

This is part of your answer, but not all of it.  The calculator will only give one answer, and will not consider answers in multiple quadrants.  We will use this answer to find any other solutions for X.

From our problem, we saw that:

cos(X) = 5/9

this means that the cosine value is positive.  In what quadrants are the cosine values positive??

Quadrant 1 and Quadrant 4 have positive cosine values.  Therefore, there will be 2 answers (one in each quadrant).

So, X = 56.25 degrees is the answer for the angle in Quadrant 1.

To get the angle in Quadrant 4 that also works, you gotta do 360 – 56.25 = 303.75

Therefore, 303.75 degrees is another answer for X.

What this is saying is that:

cos(56.25)= 5/9 and cos(303.75) = 5/9 also.

So, X = 56.25 and 303.75

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Simplifying Trig Expressions Using Identities

Q:  Simplify cos4t – sin4t

A:

First thing we gotta notice is that this is a difference of squares.  cos4t is the same as (cos2t)2.  So, we need to use what we know about factoring from algebra to factor this:

cos4t – sin4t can be factored using the difference of squares formula/concept (this just takes practice to see this and realize it):

(cos2t – sin2t)(cos2t + sin2t)

But, now we can use a trig identity because cos2t + sin2t = 1, so plug that in:

(cos2t – sin2t)(cos2t + sin2t)

(cos2t – sin2t)(1) = (cos2t – sin2t)

Now we gotta notice that even though it is simplified a bunch, we have another identity:

(cos2t – sin2t) = cos(2t)

And now we are as simplified as we get.

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Amplitude, Phase Shift and Period

Q:  Consider the function y = -5 cos (2x – .1π)

a)  Identify the amplitude

b)  Identify the phase shift

c)  Identify the period

Answer:

The following formulas / concepts will work for sin and cos graphs:

y = a cos (b*(x – h) ) + k

Note:  a, b, h and k are just numbers that will affect the graph.

The amplitude is |a|

The phase shift if 2π / |b|

The phase shift is “h”

The vertical shift is “k”

A big thing to notice:  the “b” value is factored out in this red formula.

So…. Let’s start with our actual example: y = -5 cos (2x – .1π)

Identify who is a, b, h, and k.  Notice, the b is not factored out, so let’s do that firt:

y = -5 cos (2x – .1π)

y = -5 cos (2*(x – .05π))

[if you multiply the 2 back through, you get the same as the original]

Now,

a = -5

b = 2

h = .05*pi

k = 0 (there is no + k at the end of the problem)

a)  The amplitude is |a| = |-5| = 5

b)  The phase shift is h = .05π (the graph is shifted .05π units to the right)

c)  The period is 2π/|b| = 2π/|2| = π

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Graphing a Quadratic Function: 2 methods

Q:  Graph the Quadratic Function:  f(x) = 2x2 – 3x + 1

Answer:

There are two main methods to do this.  I will do both methods (depending on what you have learned in class, you will want to pick either method 1 or method 2 that I show).

Method 1:  Plotting Points

Because this is a quadratic equation, we know the graph will be in the shape of a parabola.  So, I am going to pick a few values for x, then find the y values.  Then, I will plot the points on the graph to make a parabola.  You can pick any values for x when you do this method.

Let’s pick x = 0.  So, plug x into the equation to find y:

2(0)2 – 3(0) + 1 = 0 – 0 + 1 = 1

Point 1:  When x = 0, y = 1:  (0, 1)


Let’s pick x = 1 now:

2(1)2 – 3(1) + 1= 2(1) – 3 + 1 = 2 – 3 + 1 = 0

Point 2:  When x = 1, y = 0:  (1, 0)


Let’s pick x = -1 now:

2(-1)2 – 3(-1) + 1= 2(1) + 3 + 1 = 2 + 3 + 1 = 6

Point 3:  When x = -1, y = 6:  (-1, 6)

Now, plot the points (0, 1) and (1, 0) and (-1, 6) on a graph.  Connect the dots to make a parabola!

Keep in mind:  You may need to get more than 3 points for better accuracy.

Method 2:  Finding the y-intercept and the vertex:

The y-intercept of any type of graph occurs when you plug in 0 for x.  So, plug in 0 for x into the equation and solve for y:

Continue reading Graphing a Quadratic Function: 2 methods

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Simplify by Rationalizing the Denominator

Q:  Simplify by rationalizing the denominator:  √8/√24

Answer:  To rationalize the denominator, multiply both top and bottom by the denominator.  So, multiply both top and bottom by √24:

√8/√24

√8*√24 / √24*√24

Simplify the top and bottom like so:

√192 / 24

Now, we need to simplify the numerator!  We do this by factoring the numerator into perfect squares.  It turns out that 192 = 64*3, so:

√192 / 24 = √64*√3 / 24

And, this simplifies to:

Continue reading Simplify by Rationalizing the Denominator

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Trig Functions and Exact Values

Q:  Let X be an angle in quadrant III such that cos(X) = -12/13.  Find the exact values of csc(X) and cot(X).

A:

This will be a little hard to explain, since I cannot currently provide a picture, but it should be manageable!  Draw a picture on your own paper and follow along with me.

Draw a right triangle and label one of the angles X (just not the right angle).  We are going to ignore the negative sign for now and then take it in to account later!  Since the cos(X) = 12/13, the side adjacent to X is 12 and the hypotenuse is 13.  Use the Pythagorean Theorem to find the missing side (the side opposite X).  You will find that the opposite side is 5.

So, Adj = 12, Opp = 5 and Hyp = 13

Now, let’s consider the negative sign:  Since X is a QIII angle, we know that only tangent and cotangent are positive values in that quadrant — all other trig functions in QIII are negative.

Now, we need to find the csc(X).  “csc” is the ratio of the hyp / opp [and, it will be negative]

So, csc(X) = -13/5

And, cot(x) is the ratio of the adj / opp [and, it will be positive].  So, cot(X) = 12/5

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Combining Fraction (involving variables)

Q:  Condense:  3/2x + x/(2x-6)

Answer:

In order to add fractions, we need to have common denominators.  Remember, in order to get common denominators, we need to multiply by something.. Just because one denominator is 2x – 6 and the other is 2x, we cannot just subtract 6 to make them match.

So, the fraction on the left needs a (2x-6) and the fraction on the right needs a (2x)…  So, you have to multiply the fraction on the left by (2x-6) and the fraction on the right needs to be multiplied by (2x).  Remember:  multiply to the top and the bottom like so:

3/2x + x/(2x-6)

3(2x-6)/2x(2x-6) + x(2x)/(2x-6)(2x)

Now, simplify each numerator, leave the denominators alone:

Continue reading Combining Fraction (involving variables)

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Solving a quadratic

Q:  Solve for n:

4n2 + 3 = 7n
 

Answer:  Since there is an n-squared term, this is a quadratic equation.  In order to solve this, we need to set the whole equation equal to 0 first (so, let’s subtract the 7n over to the left side of the equation):

We get:

4n2 – 7n + 3 = 0  [notice that I put the n’s in order of n-squared, n, and then the constant 3]

Now, there are a few methods you may have learned to can help you solve this:  1)  Factoring or 2) Quadratic Formula or 3) Completing the Square.

Continue reading Solving a quadratic

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Simplifying Radicals

Q:  Simplify: √(9) / √(18)

Answer:

First, we gotta know that we can break square roots apart into their factors.  So, √(18) can be broken up into √(3)*√(6) since 3*6 is 18… Or, √(18) can be broken up into √(2)*√(9) since 2*9 is 18.

So, I am going to break √(18) = √(2)*√(9) since our problem already has a √(9) in it (and since √(9) is a number we know!).

√(9) / √(18) = √(9) / [√(2)*√(9)]

Now, there is a √(9) on top and on bottom, so it can cancel out to leave:

1 / √(2)

However, depending on what class you are in and your teacher, you may need to rationalize the denominator.  Rationalizing the denominator means to get all square roots out of the denominator and into the numerator only.

Continue reading Simplifying Radicals

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Mixture Problems

Q:  How many mL of a 3% acid solution should be mixed with a 7% acid solution to create 120 mL of a 5% acid solution?

A:

There is a generic formula we can use for mixture problems:

(quantity of solution A)*(% of A) + (quantity of solution B)*(% of B) = (final quantity)*(% of final)

So, for this problem, we know the final quantity is 120 mL.

Let’s use “x” to represent the amount of solution A that we need.  Since we need 120 total, we know that the amount of solution B can be represented by “120 – x” (the left-over).

So, plugging the values and variables into the equation:

Continue reading Mixture Problems