instantaneous rates y=14,000(1.06)^X

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fred34

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using the equation y= 14,000(1.06)^X where x is the number of years
first find the instantenous rate at x=o....then find the x value to which the instantaneous rate is twice the rate at x=0


I am really lost on this question. If someone could just show the steps so I can use this as a basis for 4 other problems I have that are just like this one.
 
what does this equation represent? Anything? Usually when you see the expression "instantaneous rate" that means find the derivative and plug in the value of x. That would be finding the instantaneous slope, but it may depend on what the numbers actually represent.
 
Sorry, I had a long post.... explaining all of this. Just before I submit though my computer shut down ( I have the windows 7 beta and haven't upgraded yet = ()

Alright... back to typing it ... again...
 
tutor_joel is correct. Finding the instantaneous rate means to take the derivative and then plug in your value of X.

Finding the derivative of this function is tricky; however, so I will give you some help with it. We have:

\(\displaystyle y = 14,000(1.06)^x\)

The first step to solve this derivative is to take the Natural Log of both sides:

\(\displaystyle ln(y) = ln(14,000(1.06)^x)\)

Now, apply a property of logarithms: \(\displaystyle ln(AB) = ln(A) + ln(B)\) :

\(\displaystyle ln(y) = ln(14,000) + ln((1.06)^x)\)

Another property of logarithms: \(\displaystyle ln(A^B) = Bln(A)\) therefore:

\(\displaystyle ln(y) = ln(14,000) + xln(1.06)\)

Now take the derivative of both sides... careful with this, you must take the derivative of ln(y) as well:

\(\displaystyle \frac{y'}{y} = ln(1.06)\)

Now, this looks a bit weird, but remember: \(\displaystyle y = 14,000(1.06)^x\), therefore:

\(\displaystyle \frac{y'}{14,000(1.06)^x} = ln(1.06)\)

Now, just solve for the derivative, y':

\(\displaystyle y' = ln(1.06)*14,000(1.06)^x\)

Another way of thinking of this derivative is by a formula, which I just derived for you:
\(\displaystyle \frac{d}{dx}A^x = ln(A)*A^x\)


From here you can get it, I believe.

Hope this helps some!
 
Okay I understand the first part now...thanks

find the x value to which the instantaneous rate is twice the rate at x=0

would you just take the value from the first part and double it then set the derivative equation equal to it and solve for x?
 
fred34 said:
Okay I understand the first part now...thanks

find the x value to which the instantaneous rate is twice the rate at x=0

would you just take the value from the first part and double it then set the derivative equation equal to it and solve for x? - Yes
Click to expand...
 
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