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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.

user1945925
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1 Answers1

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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.

Ross Millikan
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  • Explain the $1.25^{t}$ expression and how you came up with this. I've never seen that before – user1945925 Jan 15 '13 at 02:04
  • @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 04:30
  • at f(0) =1 I wasn't able to find a valid angle – user1945925 Jan 15 '13 at 07:30
  • @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