Question

For the following exercises, find $dy/dx$ at the value of the parameter.$x=4\cos (2\pi s),y=3\sin (2\pi s),s=-\frac{1}{4}$

Short answer

$\frac{dy}{dx}=0$ at $s=-\frac{1}{4}$.

Step-by-step solution

Differentiate x with respect to s

Given the function for $x$, differentiate it with respect to $s$:$$x=4\cos (2\pi s)$$Differentiating with respect to $s$, we have:$$\frac{dx}{ds}=\frac{d}{ds}[4\cos (2\pi s)]=4\cdot (-\sin (2\pi s))\cdot 2\pi =-8\pi \sin (2\pi s)$$

Differentiate y with respect to s

Given the function for $y$, differentiate it with respect to $s$:$$y=3\sin (2\pi s)$$Differentiating with respect to $s$, we have:$$\frac{dy}{ds}=\frac{d}{ds}[3\sin (2\pi s)]=3\cdot \cos (2\pi s)\cdot 2\pi =6\pi \cos (2\pi s)$$

Use the chain rule to find $\frac{dy}{dx}$

The chain rule states that $\frac{dy}{dx}=\frac{dy}{ds}\cdot \frac{ds}{dx}$. First, recall that $\frac{ds}{dx}=\frac{1}{\frac{dx}{ds}}$:$$\frac{ds}{dx}=\frac{1}{-8\pi \sin (2\pi s)}$$Now calculate $\frac{dy}{dx}$:$$\frac{dy}{dx}=6\pi \cos (2\pi s)\cdot \frac{1}{-8\pi \sin (2\pi s)}=\frac{6\pi \cos (2\pi s)}{-8\pi \sin (2\pi s)}=\frac{-3\cos (2\pi s)}{4\sin (2\pi s)}$$

Substitute s=-1/4 into $\frac{dy}{dx}$

Substitute $s=-\frac{1}{4}$ into the expression:$$\frac{dy}{dx}=\frac{-3\cos (2\pi (-\frac{1}{4}))}{4\sin (2\pi (-\frac{1}{4}))}=\frac{-3\cos (-\frac{\pi }{2})}{4\sin (-\frac{\pi }{2})}$$Evaluate $\cos (-\frac{\pi }{2})=0$ and $\sin (-\frac{\pi }{2})=-1$:$$\frac{dy}{dx}=\frac{-3(0)}{4(-1)}=0$$

Conclusion

After evaluating all expressions and substituting the given parameters, we find that the derivative of $y$ with respect to $x$ at $s=-\frac{1}{4}$ is 0.

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus, especially when dealing with composite functions. To put it simply, the chain rule helps us differentiate a function that is nested inside another function.
This can be visualized with the expression:$$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$$This formula tells us to first find the derivative of the outer function and multiply it by the derivative of the inner function.

• In our problem, we used the chain rule to connect the differentiation of functions with respect to a parameter, $s$.
• After differentiating both $x$ and $y$ with respect to $s$, we applied the chain rule to link these derivatives with $\frac{dy}{dx}$.
• We realized that to find $\frac{dy}{dx}$, we need $\frac{dy}{ds}$ and $\frac{dx}{ds}$.

In many real-world scenarios, the chain rule allows us to tackle complex differentiation in a step-by-step manner. By breaking down functions into simpler components, we can handle otherwise challenging mathematical problems more effectively.

Parametric Equations

Parametric equations represent a slightly different way of plotting curves or paths. Instead of defining $y$ directly as a function of $x$, both $x$ and $y$ are expressed in terms of a third variable, called a parameter.
This parameter, often denoted as $t$ or $s$, acts like a timeline for the path. For example:

• $x=4\cos (2\pi s)$
• $y=3\sin (2\pi s)$

In these equations:

• $s$ determines points on the curve; as $s$ changes, $x$ and $y$ describe a path.
• This technique is incredibly useful for describing circular or elliptical paths, where a single function $y=f(x)$ might not suffice.

Using parametric equations, complex shapes and motions become much more straightforward to examine, especially in physics and engineering disciplines. It allows us to represent dynamic systems where two variables change over time relative to an underlying parameter.

Differentiation with Respect to a Parameter

Differentiation with respect to a parameter involves taking the derivative of a function that is formulated in terms of a parameter instead of a single independent variable. This is crucial when dealing with parametric equations.
Here's how it works:

• For a parametric equation like $x=4\cos (2\pi s)$, differentiating with respect to the parameter $s$ involves finding $\frac{dx}{ds}$.
• Similarly, for $y=3\sin (2\pi s)$, we calculate $\frac{dy}{ds}$.

These derivatives give us the rates at which $x$ and $y$ change as the parameter $s$ changes. By analyzing these rates, we can gain insights into the behavior of the curve or path represented by the parametric equations.In practice, differentiating with respect to a parameter is vital for determining important characteristics of the path described by parametric equations, such as direction and speed of motion.