About Six Months Ago I figured something out that was very exciting to me. Let me share it with you.
Suppose you have a Certificate of Deposit that is earning a fixed interest rate for a specific term that is compounding daily. Who cares what the term is. Say you want to calculate the exact balance on a specific day. Well I will tell you how you can do it. Of course this will be assuming that it is not a leap year. For leap year calculations just change all of the 365 values in each formula to 366.
Let's first assign variables:
Let P = (The initial amount), r = (The rate), 365 = (365 days in the year), D = (The number of days that our account has been compounding) and I = (The Interest that we will get) and W = (The initial amount + the Interest we get)
The formulas are as follows:

APY (Annual Percentage Yield) = (1+ r/365)^(365) - 1

RATE = ln(APY+1) = [ln(1+r/365)^(365)] = 365*ln(1+r/365) = r

I = P*(1+r/365)^(D) - P So W = (P+I) = P*(1+r/365)^D

D = [ln(W)-ln(P)] / [ln(1+r/365)] where W = (P+I)

So how does this work? Say that we want to know how much interest we would get after 10 days of daily compounding at a 7% rate, with an initial deposit of $100,000.00 to our Certificate of Deposit. Then
I = P*(1+r/365)^(D) - P
So I = $100,000.00*(1+.07/365)^(10) - $100,000.00 = $191.95

Now what if we wanted to know how many days it would take to double our money
Doubling our money would mean that we would have to earn $100,000.00 in interest.
This means we want to know how many days will it take with an initial deposit of $100,000.00 at daily compounding at a rate of 7% to grow to be $200,000.00
So we have D = ln(W)-ln(P)/ln(1+r/365) so we have
ln($200,000.00)-ln($100,000.00)/[ln(1+.07/365)] = 3614.61 days. This means that it would take 3614.61 days to double our money at a rate of 7%

We know that the annual percentage yield is the actual rate of return on an initial deposit that has been compounding at a fixed rate for a full year.

But what if we want to calculate the actual rate of return on any given day after we make the initial deposit?
Well that is easy too. We can take calculate the interest up until that day, and then divide it by our principle account. For example we calculated that we after 10 days on a deposit of $100,000.00 we received 191.95 in interest. Therefore if we wanted to calculate the actual rate of return we would be = to interest / principle = 191.95/100,000 = .0019195 or .19% As you can see although we may be getting 7% as our rate, this rate is actually a yearly rate, in other words, our rate of return actually starts at 0%

Water is leaking out of an inverted conical tank at a rate of 10,000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has a height of 6m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank?

Hmm… since we are dealing with a conical shape the Volume (V) should = 1/3*pi*r^2*h
And the Area should = pi*r*sqr(r^2+h^2)

We are given dh/dt = 20 dV/dt = C – 10,000 where V = 1/3*pi*r^2*h at time t
A picture will help us figure this out, review my picture below:



r = 2 m
let h = 2 m
height = 6 m
The Diameter is 4 m
1) We have V = 1/3*pi*r^2*h
2) By similar triangles, h/6 = r/2 so
h = r/2 * 6 = 6r/2 = 3r and
r = 2h/6 = 1/3h
4) So we have V = 1/3*pi*(1/3*h)^(2)*h = 1/3*pi*(1/3*h)^3 = pi/27*h^3
5) So dV/dt = pi/9*h^2 *dh/dt
We were given that
6) dV/dt = pi/9*h^2 *dh/dt
We are given that dh/dt = 20
7) So we have dV/dt = C – 10,000 = pi/9*h^2*dh/dt
8) All of these figures are in cm so where as our h value = 2 is in meters so we must convert 2 meters to centimeters = 200 cm
9) Now we can plug in our values h = 200,and dh/dt = 20 to solve for C
C-10,000 = pi/9*(200)^2*(20)
C =10,000 + pi/9*(200)^2*(20) = 289252.68 cm^3/min
Therefore the constant rate at which the water is pored in to the tank is 289,252.68cm^3/min

Todays Math Problem (RELATED RATES)

Boyles’ Law states that when a sample of gas is compressed at a constant temperature, the pressure P and the volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 600 cm^3, the pressure is 150 kPa, and the pressure is increasing at a rate of 20 kPa/min. At what rate is the volume decreasing at this instant?

Hmm… we are given that PV = C, where C is a constant, So we can plug our values in for P and V where P = 150 kPa and V = 600 cm^3 therefore we get that 150*600 = C
But we are not asked to find what C equals we are asked to find out how fast the rate of V is decreasing over time. When dP/dt = 20 kPa/min
To figure this one out we need to take the derivatives of our equations using the rules for related rates. Which say that dV/dP = dV/dt*dt/dP so dV/dt = dV/dP*dP/dt and dP/dt = dP/dV*dV/dt so we get dV/dt*dP/dt = dV/dP*dP/dt*dP/dV*dV/dt = dV/dt * dP/dt so since everything else cancels out we will us dV/dt which reads the derivative of Volume in respect to time, and dP/dt which reads the derivative of Pressure in respect to time.

So lets see what we can find out.

Differentiating both sides of the equation PV = C we get P*dV/dt + V*dP/dt = 0 (by the product rule)
Subtracting V*dP/dt from both sides we get P*dV/dt = -V*dP/dt
Dividing P from both sides we get dV/dt = -V/P*dP/dt

Now we can plug in our known values, V = 600 P = 150 and dP/dt = 20
To get dV/dt = -600/150*(20) = -80

Remembering that we are asked to find out what the instantaneous rate of change is which is (dV/dt) when the instantaneous rate of change of pressure is 20 kPa/min which is (Pv/dt) and knowing that the Volume is cm^3 we get a final answer that tells us that as the pressure approaches 20 kPa/min the volume be decreasing at 80 cm^3/min.

Evaluating limits

Some formulas are difficult to evaluate their limits, however, using L’Hospital’s Rule which says that with functions in the form of. 0/0 , (infinity)/(infinity) 0^0, 1^infinity.
We can simply simplify the quotient and take the derivative until our value will work in the remaining function.

Lets take a look at some examples:

Example 1: Suppose we want to take the limit of (x^2-16)/(x-4) as x approaches 4.
If we simply plug in 4 into the function we get (4^2-16)/(4-4) = 0/0
So L’Hospital’s Rule says that we can take the derivative d/dx of the numerator and the derivative of the denominator so lets try that:
Taking d/dx(x^2-16)/d/dx(x-4) we get (2x)/1
Now we may plug in 4 into our new function to get that the limit as x approaches 4 or (x^2-16)/(x-4) = (2)*(4)/1 = 8

Lets take a look at another example:

Example 2: Suppose we want to take the limit of (4x^2-5x)/(1-3x^2) as x approaches infinity
The problem here is that infinity is not a number therefore we can not plug it into the original formula. So lets try take the derivative of the numerator and the derivative of the denominator to see what we yield.
d/dx(4x^2-5x)/d/dx(1-3x^2) = (8x-5)/(6x) we still cannot plug anything into this equations so we will again take the derivate of the numerator and the denominator of
(8x-5)/(6x), lets see d/dx(8x)/d/dx(6x) = 8/6 = 4/3 giving us that the limit as x approaches infinity of (4x^2-5x)/(1-3x^2) is 4/3

Lets do more

Example 3: Suppose we are ask to find the limit of sin(x)/x as x approaches 0 if we plugged 0 into sin(x)/x we get 0/0. so lets take the derivative of the numerator and the denominator of sin(x)/x Doing so we get d/dx(sin(x))/d/dx(x) = cos(x)/1
Now we may plug in 0 into cos(x)/1 to cos(0)/1 = 1/1
So the limits as x approaches 0 of sin(x)/x = 1

Lets try a trickier one:

Example 4: Suppose that we want to find the limit of e^x/x^2 as x approaches affinity. If x is approaching positive infinity then we have that the numerator and the denominator are both approaching affinity. Lets try to apply L’Hospital’s Rule and take the derivative of the numerator and of the denominator and see what happens.
Taking d/dx(e^x)/d/dx(x^2) we get (e^x)/(2x) so this tell us that e^x is increasing faster than 2x so this limit should be infinity, lets check by applying L’Hospital’s Rule again to (e^x)/(2x)
We get that d/dx(e^x)/d/dx(2x) = e^x/2 Now we can see that as e^x increases then 2 will go into it more times to give us the limit as x approaches infinity of e^x/x^2 = infinity

Example 5: Suppose that we want to find the limit of (5t^4-4t^2-1)/(10-t-9t^3) as x approaches 1
More numbers but don’t worry with what we know those will all go away.
Lets apply L’Hospital’s Rule and take the derivative of the numerator and the dominator multiple times to get that d/dx(5t^4-4t^2-1)/d/dx(10-t-9t^3) = (20t^3-8t)/(-1-27t^2)
we could plug in 1 into (20t^3-8t)/(-1-27t^2)
To get 20-8/(-1-27) = 12/(-28) = -6/14 = -3/7 so the limit of (5t^4-4t^2-1)/(10-t-9t^3) as x approaches 1 = -3/7

Lets try applying L’Hospital’s Rule other types of limits:

Example 6: Suppose that we want to find the limit of xlnx as x approaches 0 This function is not in quotient form but we can put it in quotient form xlnx = (lnx)/(1/x)
Now lets take the derivative the numerator and the denominator to get
d/dx(lnx)/d/dx(1/x) = (1/x)/((x*0)-1(1)/(x^2)) simplifying we get (1/x)/(-1/x^2) = x^2/-x = (-x) now we can plug in 0 into (-x) to get (-0) = 0
Therefore the limit as x approaches 0 of xlnx = 0

Lets try this one:

Example 7: Find the limit of xe^x as x approaches negative infinity
Since this is not in quotient form we can apply L’Hospital’s Rule until it is, so lets put it in quotient form. xe^x = (e^x)/(1/x) or xe^x = (x)/(1/e^x)
Now lets apply L’Hospital’s Rule on (e^x)/(1/x) d/dx(e^x)/d/dx(1/x) = (e^x)/(-1/x^2) so that is not going to help us. What if we apply L’Hospital’s Rule to (x)/(1/e^x) then we get d/dx(x)/d/dx(1/e^x) = 1/(e^x*0)-(e^x*1)/(e^x)^2 = (1)/(-e^x)/(e^x)^2 simplifying we get 1/(-e^x) therefore as x approaches –infinity the denominator is getting bigger and which means that 1/(-e^x) will go to 0, so the limit as x approaches negative infinity
of xe^x = 0

Ok one more

Example 8: This one is tricky! Find the limit of x^(1/x) as x approaches infinity it looks like it will be zero but don’t let the eyes deceive you because as x if x = infinity then (1/x) will go to zero witch means that we will have (infinity)^0 which any number x^0 = 1 so the limit of x^(1/x) should be 1
Lets check:
Let y = x(1/x)
Using our friend (ln) we can attach natural log (ln) to both side of the equation to give us
lny = lnx^(1/x) so lny = 1/x(lnx)
now we can see that as x get larger then 1/x will go to 0 which would make 1/x(lnx) = 0
however because we attached natural log to y we must get rid of the natural log by raising the equation lny = 1/x(lnx) by (e) so lets do that. Since we found that as x goes to infinity in 1/x(lnx) then 1/x(lnx) = 0 we can say e^(lny) = e^0 giving us y = 1
Therefore the limit as x approaches infinity of x^(1/x) = 1

Finding the limit

Limit as x approaches -1 of (x^2-1)/(x+1)
Solution: if x = -1 then the denominator would equal 0 which is undefined.
Therefore we are interested in finding out what the follow of (x^2-1)/(x+1) is as x gets close to -1
We can do this by taking the derivative of both the top and bottom, then plug in -1 and evaluate.

d/dx (x^2-1)/d/dx (x+1) = 2x/1; therefore the limit as x approaches -1 of (x^2-1)/(x+1) is 2(-1)/1 = -2

As illustrated in the graph below when x gets closer to -1 then the value of (x^2-1)/(x+1) gets closer to -2 therefore we know that (x^2-1)/(x+1) is always increasing.



Limit as x approaches 1 of (x^9-1)/(x^5-1)
Solution:
Step one: find the d/dx(x^9-1)/d/dx(x^5-1) = 9x^8/5x^4
Step two: evaluate 9x^8/5x^4 when x =1 =9/5
Therefore the limit as x approaches 1 of (x^9-1)/(x^5-1) = 9/5 = 1.80

As illustrated in the graph below when x gets closer to -1 then the value
Of (x^9-1)/(x^5-1) gets closer to 1 the value of (x^9-1)/(x^5-1) gets closer to 1.80




Limit as t approaches 0 of (e^t-1)/(t^3)
Solution:
Step one:
This one is a little more trickier and requires more than the previous ones because if we find d/dx(e^t-1)/d/dx(t^3) we get e^t/3t^2 and we still can’t evaluate x = 0 because the donator will still = 0 and that is undefined so we proceed with the following:
However we can see that if as t gets closer to 0 e^t gets closer to 1 and as t gets closer to 0 then 3t^2 gets closer to 0 therefore e^t/3t^2 goes to infinity and so the Limit as t approaches 0 of (e^t-1)/(t^3) is infinity as illustrated in the graph below:

Graph of t approaches 0 from the positive side



Graph of t approaches 0 from the from the negitive side

1.) if F(X) = 3x^2 -12x +5 find the absolute minimum and maximum values in the interval [0,3]
dF/dx = 6x-12
Find where 6x -12 = 0
x = 12/6 = 2
evaluate F(2) = 3(2)^2-12(2)+5 = 12-24+5 = -7
Therefore the minimum value of F on the interval [0,3] is (2,-7)
evaluate F(0) = 5 ==> (0,5)
evalute F(3) = 3(3)^2-12(3)+5 = -4 ==> (3,-4)
with a minnimum value at (2,-7)

2.) if G(X) = x^3 -3x + 1 find the absolute minimum and maximum values in the the interval [0,3]
dG/dx = 3x^2-3
Find where 3x^2-3 = 0
x = +- sqrt3/3 = +- 1
But we are ask to evaluate the interval [0,3] so we discard x = -1
evalute G(1) = (1)^3-3(1) + 1 = 1-3+ 1 = -1
Therefore the minimum value of G on the interval [0,3] is (1.-1)
evaluate G(0) = 1
evaluate G(3) = 27-9+1= 19
with a minnimum value at (1,-1)

3.) A farmer wants to fence an area of 1.5 million square feet in a rectangular field and then divide it in half with a fence parallel to one of the sideds of the rectangle. How can he do this so as to minimize the cost of the fences?

Since we are dealing with a rectangle we know that the area of a rectangle is equal to x side * y side, so x*y = 1.5 mil so x*y = 10^6 *1.5

And so y = (10^6*1.5) /x
Since we are cutting the rectangle in half we have the following picture:

We can see that the parimiter of the rectangel is going to be 3x+2y
and we found that y = (10^6*1.5) /x
so we can subtitute y into 3x+3y = 3x+2*(10^6*1.5 /x) = 3x+3*10^6/x
finding the Derivative of 3x+3*10^6/x
we get that dy/dx = 3-3*10^6/(x)^2 = 3(x^2-10^6)/(x^2)
The minimum will be found when (X^2-10^6) = 0
which occures when x = 10^3
we evaluate x = 10^3 in the origional equation y= (10^6*1.5) /x
y = (10^6*1.5)/(10^3) = 1500
Therefore we found that in order to minimize the cost of the fence the farmer should build the fence to be 1000 by 1500. That all x sides should = 1000 ft. and all y sides should equal 1500 ft.