Differentiating Complex Exponentials

[Matty R]

Hello

I'm currently using past papers to revise for January exams, and I've found a bit of a problem with something I thought I was okay with.

Homework Statement

The position at time t of a particle undergoing damped oscillations is given by:

$$x = 2e^{-t}\sin(3t)$$.

Express this in terms of a single complex exponential.

Hence evaluate the particle's velocity, $$v = \frac{dy}{dx}$$

Homework Equations

$$e^{i \theta} = cos(\theta) + isin(\theta)$$

Chain Rule: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

The Attempt at a Solution

$$x = 2e^{-t}sin(3t)$$

$$sin(3t) = \text{Im} \left[e^{i3t} \right]$$

$$x = \text{Im} \left[2e^{-t}\cdot e^{i3t} \right]$$

$$= 2 \text{Im} \left[e^{-t} \cdot e^{i3t} \right]$$

$$= 2 \text{Im} \left[e^{-t + i3t} \right]$$

$$= 2 \text{Im} \left[ e^{(-1 + 3i)t} \right]$$

$$v = \frac{dx}{dt}$$

$$= 2 \text{Im} \left[(-1 + 3i) e^{-t} (cos(3t) + isin(3t)) \right]$$

$$= 2e^{-t} (-cos(3t) - 3sin(3t))$$

Thats the answer I get by following the only example I have of this from my lecture notes. However, that example was concerned with the real part of the complex exponential.

The baseline solutions to the paper give the following answer:

$$v = 2e^{-t}(3cos(3t) - sin(3t))$$

Does anyone know what I've done wrong, or missed?

Thanks.

 

[tiny-tim]

Hello Matty R!

You've taken the real part …

just take the other bits instead! ​

(and chuck away the i)

 

[Matty R]

Hiya tim. Nice to see you again.

That makes perfect sense. I've managed to get the same answer as the solutions now, and I was wondering if you'd be so kind as to confirm what I did.

For the real part:

$$2e^{-t} \left[(-1 \times cos(3t)) + (3i \times isin(3t) \right]$$

For the imaginary part:

$$2e^{-t} \left[(3i \times cos(3t)) + (-1 \times isin(3t) \right]$$

Then discard the "$$i$$"s

 

[tiny-tim]

Yup!

 

[Matty R]

Brilliant. Thank you very much.