Not sure what to do (rate of change of % of public aware)  TOPIC_SOLVED

Limits, differentiation, related rates, integration, trig integrals, etc.

Not sure what to do (rate of change of % of public aware)

Postby stackey82 on Fri Mar 27, 2009 9:08 pm

I have a problem that I believe I got but want to make sure. The question follows-

The awareness of the product is appx. by A(t)= (10t^2)(2^-t), where t is the time in months. Find the rate of change of the percent of public that is aware of the product after 5 months.

I used the product rule but not sure if I applied it properly
(20t)(2^-t)+(10t^2)(2t^(-t-1))
Then combine and this is where i think i messed up
((20t)/(2^t))+((10t^2)/(2t^(t+1))
Then I plugged in the 5 for t and got 3.3125 but this doesn't seem right to me. If I am right great if not can someone show me where I went wrong? Thanks and it would be greatly appreciated.
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Postby stapel_eliz on Fri Mar 27, 2009 9:14 pm

stackey82 wrote:The awareness of the product is appx. by A(t)= (10t^2)(2^-t), where t is the time in months. Find the rate of change of the percent of public that is aware of the product after 5 months.

I used the product rule but not sure if I applied it properly
(20t)(2^-t)+(10t^2)(2t^(-t-1))

You're find on the derivative of the 10t2 part, but you might want to review the rules for exponentials. They are not the same as polynomials! :oops:

In general, for f(x) = ax, the derivative is f'(x) = ax ln(a), where "ln(a)" is the natural logarithm of "a".

In your case, a = 2. :wink:
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Re: Not sure what to do (rate of change of % of public aware)  TOPIC_SOLVED

Postby stackey82 on Sat Mar 28, 2009 1:02 am

Original problem (10t^2)(2^-t)

So I would get
(20t)(2^-t)+(ln2)(2^-t)(-1)(10t^2) with t=5

3.125+(-5.415)= -2.29
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Postby stapel_eliz on Sat Mar 28, 2009 2:57 pm

I get the same result. Good work! :thumb:
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