Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Determine an expression for an exponentially growing sinusoid that oscillates five time per second, whose amplitude envelope increases 25% every second, and whose amplitude at $t = 0$ is $1$, and whose derivative at $t = 0$ is 54.6371.

What I have done: For the sinusoid portion, I know that I need a frequency of 5Hz so that it oscillates 5 times every second. Thus, this part will be $cos(2 \pi 5t)$.

I need the amplitude envelope to increase 25% every second. In decimal form 25% = 0.25
Then the derivative of a general growing exponential equating it to what we want, $ae^{at} = 0.25e^{at} $ then we solve $a=0.25$.
Hence $e^{0.25t}$.

So my final expression becomes $e^{0.25t}cos(2 \pi 5t)$.

Where at $t = 0$ the amplitude is $1$ as it should be, but the derivative at $t = 0$ is 0.25 when it should be 54.6371.

I cant figure out where I am going wrong. Thanks.

share|cite|improve this question
up vote 0 down vote accepted

First, if you want the amplitude to increase by $25\%$ every second, you want the envelope to be $1.25^t$, which equates to $e^{t\log 1.25}\approx e^{0.223 t}$, not too far off. Second, you have freedom of the phase at $t=0$. When you say the amplitude at $t=0$ is $1$, do you mean the function value or the multiplier of the sinusoid? If you mean the multiplier of the sinusoid, I can't do better than $e^{0.223 t}\sin (10 \pi t)$ with derivative $10 \pi$ at $t=0$, but the value is $0$ at $t=0$. If you mean f(0)=1, you should be able to find a phase in between that makes the function value $1$ at $t=0$ and gives the derivative you want.

share|cite|improve this answer
Explain the $1.25^{t}$ expression and how you came up with this. I've never seen that before – user1945925 Jan 15 '13 at 2:04
@user1945925: If you want something to grow by 25%, you multiply it by 1.25. If you want it to grow by 25% each second, you multiply it by 1.25 for each second. Taking the natural log gives you the same function-that is the definition of log. The reason the exponent is less than 0.25 is essentially the same as compound interest: $e^x=1+x+x^2/2!+\ldots$. For $x \ll 1$ the later terms don't contribute much, but they do some. – Ross Millikan Jan 15 '13 at 4:30
at f(0) =1 I wasn't able to find a valid angle – user1945925 Jan 15 '13 at 7:30
@user1945925: The derivative of $e^{0.223t}\cos 10t$ is $0.223$ at $t=0$ and the derivative of $e^{0.223t}\sin 10t$ is $10 \pi$. So we can use the function $e^{0.223t}\(cos (10 \pi t) + \frac {54.6371-0.223}{10 \pi} \sin (10 \pi t))$. The first term gives $f(0)=1$, the second makes up the derivative shortfall. – Ross Millikan Jan 15 '13 at 14:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.