Question

In each of the following problems, $z$ represents the displacement of a particle from the origin. Find (as functions of $t$ ) its speed and the magnitude of its acceleration, and describe the motion. $z=(1+i)e^{it}$.

Short answer

The speed is $\sqrt{2}$, the magnitude of the acceleration is $\sqrt{2}$, and the particle exhibits uniform circular motion.

Step-by-step solution

Express the given function in terms of real and imaginary parts

The displacement of a particle is given by the complex function equation: $z=(1+i)e^{it}$. Rewrite the complex exponential in terms of cosine and sine: $e^{it}=\cos (t)+i\sin (t)$. Thus, $z=(1+i)(\cos (t)+i\sin (t))$.

Distribute the terms

Multiply out the terms:$$z=(1+i)(\cos (t)+i\sin (t))$$$$z=\cos (t)+i\cos (t)+i\sin (t)-\sin (t)$$$$z=\cos (t)-\sin (t)+i(\cos (t)+\sin (t))$$. Therefore, the real part is $\cos (t)-\sin (t)$ and the imaginary part is $\cos (t)+\sin (t)$.

Find the velocity

The speed of the particle is the magnitude of the velocity vector. First, find the velocity by differentiating each component with respect to $t$:$$v=\frac{dz}{dt}$$For the real part: $\frac{d}{dt}(\cos (t)-\sin (t))=-\sin (t)-\cos (t)$For the imaginary part: $\frac{d}{dt}(\cos (t)+\sin (t))=-\sin (t)+\cos (t)$Thus, the velocity is $v=(-\sin (t)-\cos (t))+i(-\sin (t)+\cos (t))$.

Calculate the speed

The speed is the magnitude of the velocity vector:$$|v|=\sqrt{(-\sin (t)-\cos (t){)}^2+(-\sin (t)+\cos (t){)}^2}$$This simplifies to: $$|v|=\sqrt{{\sin }^2(t)+2\sin (t)\cos (t)+{\cos }^2(t)+{\sin }^2(t)-2\sin (t)\cos (t)+{\cos }^2(t)}$$$$|v|=\sqrt{2({\cos }^2(t)+{\sin }^2(t))}$$Since ${\cos }^2(t)+{\sin }^2(t)=1$, we have: $|v|=\sqrt{2}$.

Find the acceleration

The magnitude of the acceleration is the magnitude of the second derivative of $z$ with respect to $t$:$$a=\frac{d^{2}z}{d{t}^2}$$Differentiating the velocity, for the real part: $\frac{d}{dt}(-\sin (t)-\cos (t))=-\cos (t)+\sin (t)$ and for the imaginary part: $\frac{d}{dt}(-\sin (t)+\cos (t))=-\cos (t)-\sin (t)$Thus, the acceleration is $a=(-\cos (t)+\sin (t))+i(-\cos (t)-\sin (t))$.

Calculate the magnitude of the acceleration

The magnitude of the acceleration is:$$|a|=\sqrt{(-\cos (t)+\sin (t){)}^2+(-\cos (t)-\sin (t){)}^2}$$This simplifies to: $$|a|=\sqrt{{\cos }^2(t)-2\cos (t)\sin (t)+{\sin }^2(t)+{\cos }^2(t)+2\cos (t)\sin (t)+{\sin }^2(t)}$$$$|a|=\sqrt{2({\cos }^2(t)+{\sin }^2(t))}$$Since ${\cos }^2(t)+{\sin }^2(t)=1$, we get: $|a|=\sqrt{2}$.

Describe the motion

The particle's speed and magnitude of acceleration are both constant $\sqrt{2}$. The motion is thus uniform circular motion with constant speed and constant centripetal acceleration.

Key concepts

Complex functions

Complex functions involve numbers that have both a real part and an imaginary part. The displacement of the particle in this problem is represented by the complex function: $z=(1+i)e^{it}$. Understanding the real and imaginary components of this function is crucial. The exponential form $e^{it}$ can be transformed into trigonometric functions using Euler's formula: $e^{it}=\cos (t)+i\sin (t)$. This allows us to manipulate the displacement function more easily.

Differentiation

Differentiation is the process of finding the rate at which a function changes at any given point. When analyzing the motion of a particle, we differentiate the displacement function to find the velocity and then differentiate again to find the acceleration. For the displacement $z=(1+i)(\cos (t)+i\sin (t))$, we differentiate the real part and the imaginary part separately to find the velocity. Similarly, to find the acceleration, we differentiate the velocity function.

Uniform circular motion

Uniform circular motion occurs when a particle moves along a circular path with constant speed. From the given problem, we find that the particle's speed and the magnitude of its acceleration are constant $\sqrt{2}$. The constant speed implies that the particle undergoes uniform circular motion, as it moves in a path with a fixed radius and speed. The centripetal acceleration is always directed towards the center of the circle, and its magnitude is also consistent.

Velocity

Velocity is a vector quantity that represents the rate of change of displacement. It has both magnitude and direction. In this problem, we find the velocity by differentiating the displacement function $z$. The resulting velocity function is $v=(-\sin (t)-\cos (t))+i(-\sin (t)+\cos (t))$. The speed of the particle is the magnitude of the velocity vector, calculated as $|v|=\sqrt{2}$.

Acceleration

Acceleration is the rate of change of velocity. It also has both magnitude and direction. We find the acceleration by differentiating the velocity function. For this problem, the acceleration is $a=(-\cos (t)+\sin (t))+i(-\cos (t)-\sin (t))$. The magnitude of the acceleration is calculated as $|a|=\sqrt{2}$, indicating that, like the velocity, it remains constant throughout the motion.

Imaginary and real parts

Complex functions can be separated into real and imaginary parts. For $z=(1+i)(\cos (t)+i\sin (t))$, the real part is $\cos (t)-\sin (t)$, and the imaginary part is $\cos (t)+\sin (t)$. This separation is essential for differentiation and understanding the particle's motion. The real and imaginary parts of the velocity and acceleration also aid in visualizing the direction and magnitude of these vectors.