Question

If \(x=r \cos \theta, y=r \sin \theta,\) show that \(y(d x / d t)-x(d y / d t)=-r^{2}(d \theta / d t)\)

Short answer

Question: Show that if \(x = r\cos\theta\) and \(y = r\sin\theta\), then \(y\frac{dx}{dt} - x\frac{dy}{dt} = -r^2\frac{d\theta}{dt}\). Answer: By applying the chain rule, finding the derivatives for x and y with respect to time, and simplifying the resulting expression, we can establish the relationship \(y\frac{dx}{dt} - x\frac{dy}{dt} = -r^2\frac{d\theta}{dt}\).

Step-by-step solution

Rewrite given equations for dx/dt, dy/dt, and dθ/dt

Using the chain rule, we can rewrite the given equations as following: \(\frac{d}{dt}(x)=\frac{d}{dt}(r\cos\theta)\) \(\frac{d}{dt}(y)=\frac{d}{dt}(r\sin\theta)\)

Find the derivatives of x and y with respect to time t

By applying the chain rule to the expressions found in step 1, we can find the derivatives of x and y: \(\frac{dx}{dt}=\frac{dr}{dt}\cos\theta-r\sin\theta\frac{d\theta}{dt}\) \(\frac{dy}{dt}=\frac{dr}{dt}\sin\theta+r\cos\theta\frac{d\theta}{dt}\)

Compute y(dx/dt) - x(dy/dt) by substituting the expressions found in step 2 wiąz

Now, we'll find the expression \(y\frac{dx}{dt} - x\frac{dy}{dt}\) by substituting our results from step 2: \(y\frac{dx}{dt} - x\frac{dy}{dt}=(r\sin\theta)(\frac{dr}{dt}\cos\theta-r\sin\theta\frac{d\theta}{dt}) -(r\cos\theta)(\frac{dr}{dt}\sin\theta+r\cos\theta\frac{d\theta}{dt})\)

Simplify the expression and show the result equals -r^2(dθ/dt)

We will now simplify the expression obtained in step 3: \(y\frac{dx}{dt} - x\frac{dy}{dt} = r\sin\theta\frac{dr}{dt}\cos\theta-r^2\sin^2\theta\frac{d\theta}{dt} - r\cos\theta\frac{dr}{dt}\sin\theta-r^2\cos^2\theta\frac{d\theta}{dt}\) Combine like terms: \(y\frac{dx}{dt} - x\frac{dy}{dt} = -r^2\sin^2\theta\frac{d\theta}{dt} - r^2\cos^2\theta\frac{d\theta}{dt}\) Now we can factor out \(-r^2\frac{d\theta}{dt}\) from the expression: \(y\frac{dx}{dt} - x\frac{dy}{dt} = -r^2\frac{d\theta}{dt}(\sin^2\theta+\cos^2\theta)\) Since \(\sin^2\theta+\cos^2\theta=1\), we finally obtain the desired result: \(y\frac{dx}{dt} - x\frac{dy}{dt} = -r^2\frac{d\theta}{dt}\)

Key concepts

The Chain Rule

The chain rule is a fundamental technique in calculus used to find the derivative of composite functions, which are functions made by combining other functions. In the context of our exercise, where we are working with two variables, time (t) and the angle (θ), chained together through polar coordinates, the chain rule becomes indispensable.

Specifically, when we want to differentiate a function like \( r \times \text{cos}(\theta) \) with respect to time, we see that both \( r \) and \( \theta \) can vary with time. The chain rule tells us how to handle this situation: To find the derivative of the composite function, we take the derivative of the outer function with the inner function untouched, then multiply it by the derivative of the inner function with respect to the variable we are differentiating. This process is applied sequentially if multiple layers of functions exist.

Application in the Exercise

When we applied the chain rule to \( r\text{cos}(θ) \), we treated \( r \) and \( θ \) as functions of time. This allowed us to differentiate like so: \( \frac{d}{dt}(r\text{cos}(θ)) = \frac{dr}{dt}\text{cos}(θ) - r\text{sin}(θ)\frac{dθ}{dt} \), considering both the change of radius \( r \) and the change of angle \( θ \) over time.

Polar Coordinates

Polar coordinates provide an alternative to the standard Cartesian coordinate system. Instead of using a grid of vertical and horizontal lines to locate points, polar coordinates use an angle and a distance from a central point, known as the pole. Every position can be determined by the radial distance \( r \) and the angle \( θ \), where the angle is measured from a reference direction, typically the positive x-axis.

In polar coordinates, the equations \( x = r\text{cos}(θ) \) and \( y = r\text{sin}(θ) \) are used to convert polar coordinates to Cartesian coordinates. This system is particularly useful in situations where natural symmetry suggests circles or spirals, which is often the case in physics and engineering problems involving periodic motion or waves.

Understanding Through the Exercise

The exercise provided a great example of the advantage of polar coordinates. By representing the point's position using \( r \) and \( θ \) instead of \( x \) and \( y \), we were able to express the relationship between linear motion and rotational motion succinctly, leading up to the solution that incorporated both radian (angle) change and linear distance.

Derivatives

Derivatives represent the rate at which one variable changes with respect to another. It is one of the most essential concepts in calculus and provides a mathematical framework for describing how a quantity varies instantaneously. In our exercise, we were interested in how the position, described by \( x \) and \( y \), changes over time, necessitating the use of derivatives.

When we talk about the derivative of a function, symbolically represented as \( \frac{dy}{dx} \) where \( y \) is a function of \( x \), we are looking at how a tiny change in \( x \) causes a change in \( y \). In the physical world, this can represent how quickly a car's position changes with time (speed) or how a material's temperature varies at different points (thermal gradient).

Connection to the Solution

The exercise demanded us to find the derivative of \( x \) and \( y \) with respect to time (\( t \)), \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), respectively. Derivatives enable us to describe these rates of change in a precise mathematical language, allowing us to manipulate and eventually reach the conclusion that the mixed derivative \( y\frac{dx}{dt} - x\frac{dy}{dt} \) equates to \( -r^2\frac{dθ}{dt} \) signifying a beautiful interplay between linear and angular velocity components in polar coordinates.