Adding Rational Expressions

Q:  Add x/(x+1) + 2x/(x+1)

also, identify any x-values for which the expression is undefined.

Answer:  In order to add rational expressions, you need common denominators.  These expressions already have a common denominator of (x+1) so the hard work is done.

You can just add the numerators together and combine the fraction (leave the denominator alone) to get:

x/(x+1) + 2x/(x+1) = (x + 2x) /(x+1) = 3x / (x + 1) [final answer]

Now, we need to identify any x-values for which the expression is undefined.  We need to know that you can never divide by 0, so the denominator of any rational expression can never equal 0.

So, look at the denominator and find where it equals 0:

x+1 = 0

Solve for x to get:

x = -1

Therefore, the denominator is 0 if x = -1.  This tells us that x can never be -1.  The problem is undefined when x = -1.

Implicit Differentiation

Q: 

A.  Implicitly find dy/dx of exy=8

B.  Now, solve exy=8 for y first, then take the derivative.  Compare your answers to A and B.

Answer:

A.  Remember, implicit differentiation is just the chain rule:  y is a function of x.  So, we need to use the product rule since we are multiplying two functions ex times y:

Also remember, the derivative of x with respect to x is 1 and the derivative of y with respect to x is dy/dx:

Differentiate as follows:

exy + ex(dy/dx) = 0

Now, solve for dy/dx:

ex(dy/dx) = -exy

dy/dx = -exy/ex

dy/dx = -y [final answer]

However, we may want to put the answer in terms of x.  If this is the case, we know from the original problem:

exy=8 which implies:

y = 8/ex

So, dy/dx = -y = -8/ex

B.  Solve for y first, then take the derivative:

We just solved for y above and got:

y = 8/ex

We can rewrite this as:

y = 8*e-x

Now, take the derivative (using the chain rule)

dy/dx = 8*e-x*(-1)

dy/dx = -8*e-x

Which can then be re-written as:

dy/dx = -8/ex

Compare A and B?  Notice they are identical — which is good news.  Whether we solve for y first or differentiate implicitly, we still get the same answer for dy/dx.

Marginal Cost (Integration)

Q:  Suppose the marginal cost is:  MC = 12 + 8x with C(10) = 1020.  Find the fixed and total cost of producing 80 units.

Answer:

We gotta know that the marginal cost is derivative of the cost.  So, to find the cost function, we need to integrate the marginal cost function:

int[12 + 8x] = 12x + 4x2 + C

So, the cost function, C(x) = 12x + 4x2 + C

And we know that C(10) = 1020.  So, we can plug in these values to solve for the constant C like so:

1020 = 12(10) + 4(10)2 + C

1020 = 120 + 400 + C

1020 = 520 + C

500 = C

So, the cost equation is:

C(x) = 12x + 4x2 + 500

The “C” value represents the “fixed costs”.  So, the fixed costs are $500.

The total cost of producing 80 units can be found by calculating C(80):

C(80) = 12(80)+4(80)^2+500 = $27060

Integration Examples

Q:

Integrate:  14*(12 – 4x)(12x – 2x2)6 dx

Answer:

Since I see that (12 – 4x) is the derivative of (12x – 2x2), my instinct tells me to use u-substitution.

So, let

u = 12x – 2x2

then, compute du:

du = 12 – 4x dx

Now, we can directly substitute in u and du into our problem:

Integrate:  14*(12 – 4x)(12x – 2x2)6 dx = 14*u6 du

[the u replaced (12x – 2x2) and the du replaced the (12 – 4x) dx]

So, we are integrating:  14*u6 du

Using the power rule:

int[ 14*u6 du] = 14/7 * u7 + C = 2u7 + C

And since u = (12x – 2x2), the final answer is:

(12x – 2x2)7 + C

Right Triangles

Question by Photo:

Answer:

1.  Length of AB:  This can be found by using the Pythagorean Theorem.  In our case,

AB^2 + AC^2 = BC^2

We know AC = 3sqrt(5) and BC = 4sqrt(5).  Plug those numbers in to solve for AB like so:

AB^2 + [3sqrt(5)]^2 = [4sqrt(5)]^2

Now, solve (the “squared” must distribute to the 3 and the sqrt(5) on the left side, and same on the right side):

AB^2 + 9*5 = 16*5

AB^2 + 45 = 80

AB^2 = 35

AB = sqrt(35)  Final answer.

2.  Find the area of triangle ABC

Use AC as the base and AB as the height.  Use the formula:  Area = 1/2 * base * height

Area = 1/2 * AC * AB

Area = 1/2 * 3sqrt(5) * sqrt(35)

Area = 3/2 * sqrt(175)  [you can multiply the 1/2 and the 3… And you can multiply the 5 by 35 inside of the square roots]

Now, you probably need to simplify the radical:

Area = 3/2 * sqrt(175) = 3/2 *sqrt(25*7) = 3/2 sqrt(25)*sqrt(7) = 3/2 * 5 * sqrt(7) = 15/2 * sqrt(7)

 

 

 

 

Implicit Differentiation Example

Q:  Find y'(3) using implicit differentiation of the equation:

2x^2 + 3x + xy = 3

and y(3) = -8

Answer:

First, take the derivative of the equation (this will involve a product rule on the “xy” term):

The implicit differentiation of 2x^2 + 3x + xy = 3 is:

4x + 3 + xy’ + y = 0

And we know that when x = 3, y = -8.  So, plug that in and the solve for y’:

4(3) + 3 + 3y’ + (-8) = 0

12 + 3 + 3y’ – 8 = 0

7 + 3y’ = 0

3y’ = -7

y’ = -7/3

So, y'(3) = -7/3

 

Half-Life

Q:  A scientist has 256 g of goo. After 195 minutes, her sample has decayed to 16 g. What’s the half-life of the goo in minutes? (Assume exponential decay)

Answer:

First we need to set up the problem with our decay formula.  Depending on your school / teacher / book, the variables in the formulas may be different letters, but they all mean the same thing:

P = A*e^(rt)

P is the ending amount, A is the initial amount, r is the growth rate/decay, and t is time.

So, the set-up with our info:

16 = 256*e^(r*195)

Now we need to solve for r:

16 = 256*e^(r*195)

1/16 = e^(r*195)

Take the natural log of both sides:

ln(1/16) = ln(e^(r*195))

Exponents can come down as multipliers (log rules):

ln(1/16) = (195r)*ln(e)

And, ln(e) = 1, so:

ln(1/16) = (195r)

ln(1/16) / 195 = r

Plug into a calculator to get:

r = -.01422

Now, we have r.  We now need to find the half-life.  So, if we started with 256 g, we want to know how long (t) it takes to get to 128 grams.  Plug the info in to our equation:

128 = 256*e^(-.01422*t)

and solve for t:

1/2 = e^(-.01422*t)

ln(1/2) = ln(e^(-.01422*t))

ln(1/2) = -.01422*t

ln(1/2)/-.01422 = t

So, t = 48.74 minutes

The half-life of the goo is approximately 48.74 minutes.

Central Limit Theorem and Confidence Intervals

Q:  Motorola wishes to estimate the mean talk times for one of their new phones before the battery must be recharged.  In a random sample of 35 phones, the sample mean talk time is 325 minutes (x-bar).

a)  Why can we say that the sampling distribution of x-bar is approximately normal?

b)  Construct a 90% confidence interval for the mean talk time of all new Motorola phones, assuming the population standard deviation is 31 minutes.

Answer:

(Please keep in mind, statistics is not 100% my area of expertise, but I will do my best to provide clear and accurate explanations)

a)  The Central Limit Theorem essential tells us that x-bar will be approximately normally distributed for a sufficiently large number of independent random variables (which we are dealing with).  So, we can assume that the distributions of the sample means (x-bars) is normally distributed thanks to the CLT.  A sample size of 30 or larger is considered “large enough” to assume an approximately normal distribution.

b)  We first need to find (or know) the z-value associated with a 90% confidence interval.  Our alpha-value is .10 (10%), which makes alpha/2 = .05.  So, we need to find the z-value that is associated with a .05 area under the curve (a reverse look-up on a standard normal table of values). The desired z-score is 1.645

In other words, of all the possible x-bar values along the horizontal axis of the normal distribution curve, 90% of them should be within a z-score of 1.645 from the population mean.

A confidence interval can be calculated by using the following formula:

x-bar +/- z*(population standard deviation / sqrt(sample size))

So, a 90% confidence interval, with our given information can be calculated by:

325 +/- 1.645*(31/sqrt(35))

325 +/- 8.6197

Which gives two answers:

325 – 8.6197 = 316.3803

and

325 + 8.6197 = 333.6197

Therefore, we can be 90% confident that the average length of talk time before needing recharge for the population is between 316.3803 and 333.6197 minutes.

Derivative and Power Rule

Q:  Find f ‘ (t) if f(t) = sqrt(3)/t^4

(using the power rule)

Answer:

Step 1)  Rewrite the problem to get t^4 out of the denominator:

sqrt(3)/t^4 = sqrt(3)*t^-4

So, now take the derivative of sqrt(3)*t^-4

Remember, “sqrt(3)” is a constant multiplier, so it just stays along for the ride.  The power rule tells you to bring the exponent down as a multiplier and then subtract 1 from the exponent, like so:

f(t) = sqrt(3)*t^-4

f ‘ (t) = sqrt(3)*(-4)*t^(-4-1)

f ‘ (t) = -4*sqrt(3)t^-5  [final answer]

You could rewrite f ‘ (t) to putt back in the denominator like so:

f ‘ (t) = -4*sqrt(3) / t^5

Solve the system by using matrices (and inverse matrices)

Q:  Solve the system of equations by using matrices (and inverse matrices)

7x + 5y = 3

3x – 2y = 22

Answer:

Okay, first set up the coefficient matrix (A), the variable matrix (X) and the “answer matrix” (B) so that:

A*X = B

| 7  5 | | x | = | 3 |

|3  -2| | y |     |22|

A       X   =   B

So, if A*X = B, and we want to solve for X, we know with matrices that:

A^-1 * A*X = A^-1 * B  (Note:  A^-1 means A inverse)

Which simplifies to:

X = A^-1 * B

So, all we need to do is find A^-1 and multiply it by B to solve for X.

To find an inverse of a 2 x 2 matrix:

If A =

| a  b |

| c  d |

then    A^-1 =

1/(detA) *

| d  -b |

| -c   a |

 

So, in our case:

A =

| 7  5 |

|3  -2|

And the det(A) = -14 – 15 = -29

So,

A^-1 =

(1/-29) *

| -2  -5 |

| -3    7|

A^-1 =

| 2/29   5/29 |

|3/29  -7/29 |

Now take A^-1 and multiply it by matrix B… You should get (didn’t show work):

A^-1*B =

| 4 |

| -5|

So, x = 4 and y = -5.