Finding Derivatives And Optimizing Cylinder Surface Area

This article delves into two distinct yet interconnected mathematical problems. The first part focuses on differential calculus, specifically finding the derivative ${\frac{dy}{dx}}$ for two given functions. This involves applying fundamental rules of differentiation, such as the product rule, quotient rule, and chain rule, to arrive at the derivatives. The second part transitions to an optimization problem related to a solid cylinder. We will explore how to express the surface area of the cylinder in terms of its radius, given a fixed volume, and then discuss how to determine the dimensions that minimize this surface area. These types of optimization problems are crucial in various fields, including engineering and manufacturing, where minimizing material usage while maintaining desired volume is a key concern.

1) Finding ${\frac{dy}{dx}}$ for Functions

In this section, we will determine the derivatives of two functions with respect to ${x}$. Understanding how to find ${\frac{dy}{dx}}$ is crucial in calculus as it represents the instantaneous rate of change of ${y}$ with respect to ${x}$. This has wide-ranging applications, from physics (velocity and acceleration) to economics (marginal cost and revenue). Finding the derivative often involves applying specific differentiation rules, such as the power rule, product rule, quotient rule, and chain rule, depending on the structure of the function. We will demonstrate the application of these rules in the following examples. These examples will showcase how to break down complex functions into simpler parts and apply the appropriate rules to find their derivatives. This process is fundamental to solving many problems in calculus and its applications.

1.1) ${y = x \sin^2(3x)}$

To find ${\frac{dy}{dx}}$ for ${y = x \sin^2(3x)}$, we will use the product rule and the chain rule. The product rule states that if ${y = uv}$, then ${\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}}$. Here, let ${u = x}$ and ${v = \sin^2(3x)}$. Applying the product rule requires us to find ${\frac{du}{dx}}$ and ${\frac{dv}{dx}}$. The derivative of ${u = x}$ with respect to ${x}$ is simply ${\frac{du}{dx} = 1}$. To find ${\frac{dv}{dx}}$, where ${v = \sin^2(3x)}$, we need to use the chain rule. The chain rule is essential when differentiating composite functions, that is, functions within functions. It states that if ${y = f(g(x))}$, then ${\frac{dy}{dx} = f'(g(x)) imes g'(x)}$. In our case, we can think of ${v = \sin^2(3x)}$ as a composite function where the outer function is squaring and the inner function is ${\sin(3x)}$. Applying the chain rule, we first differentiate the outer function (squaring) with respect to the inner function, then multiply by the derivative of the inner function. Thus, we have

\begin{aligned} \frac{dv}{dx} &= 2\sin(3x) imes \frac{d}{dx}(\sin(3x)) \ &= 2\sin(3x) imes \cos(3x) imes \frac{d}{dx}(3x) \ &= 2\sin(3x) \cos(3x) imes 3 \ &= 6\sin(3x)\cos(3x) \end{aligned}

Now, we can apply the product rule:

\begin{aligned} \frac{dy}{dx} &= u\frac{dv}{dx} + v\frac{du}{dx} \ &= x(6\sin(3x)\cos(3x)) + \sin^2(3x)(1) \ &= 6x\sin(3x)\cos(3x) + \sin^2(3x) \end{aligned}

We can further simplify this expression using the trigonometric identity ${2\sin(\theta)\cos(\theta) = \sin(2\theta)}$:

\begin{aligned} \frac{dy}{dx} &= 3x(2\sin(3x)\cos(3x)) + \sin^2(3x) \ &= 3x\sin(6x) + \sin^2(3x) \end{aligned}

Therefore, the derivative of ${y = x \sin^2(3x)}$ with respect to ${x}$ is ${\frac{dy}{dx} = 3x\sin(6x) + \sin^2(3x)}$. This result demonstrates the combined application of the product rule and the chain rule, which are fundamental techniques in differential calculus. Understanding these rules is crucial for differentiating a wide range of functions, especially those involving trigonometric functions and composite functions. The ability to correctly apply these rules is essential for solving many problems in calculus and its applications.

1.2) ${y = \frac{e^{2x}}{x+1}}$

To find ${\frac{dy}{dx}}$ for ${y = \frac{e^{2x}}{x+1}}$, we will use the quotient rule. The quotient rule is specifically designed for differentiating functions that are expressed as a ratio of two other functions. It states that if ${y = \frac{u}{v}}$, then ${\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}}$. In this case, let ${u = e^{2x}}$ and ${v = x+1}$. Applying the quotient rule requires us to find ${\frac{du}{dx}}$ and ${\frac{dv}{dx}}$. The derivative of ${u = e^{2x}}$ with respect to ${x}$ requires the chain rule. The chain rule, as discussed previously, is essential for differentiating composite functions. Here, the outer function is the exponential function and the inner function is ${2x}$. Applying the chain rule, we get:

\begin{aligned} \frac{du}{dx} &= \frac{d}{dx}(e^{2x}) \ &= e^{2x} imes \frac{d}{dx}(2x) \ &= e^{2x} imes 2 \ &= 2e^{2x} \end{aligned}

The derivative of ${v = x+1}$ with respect to ${x}$ is simply ${\frac{dv}{dx} = 1}$. Now, we can apply the quotient rule:

\begin{aligned} \frac{dy}{dx} &= \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \ &= \frac{(x+1)(2e^{2x}) - e^{{2x}(1)}{(x+1)}2} \ &= \frac{2xe^{2x} + 2e^{2x} - e^{2x}}{(x+1)2} \ &= \frac{2xe^{2x} + e^{2x}}{(x+1)2} \ &= \frac{e^{2x}(2x + 1)}{(x+1)^2} \end{aligned}

Therefore, the derivative of ${y = \frac{e^{2x}}{x+1}}$ with respect to ${x}$ is ${\frac{dy}{dx} = \frac{e^{2x}(2x + 1)}{(x+1)^2}}$. This result highlights the application of the quotient rule in conjunction with the chain rule. Mastering the quotient rule is crucial for differentiating rational functions, where one function is divided by another. The ability to correctly apply this rule, along with other differentiation techniques, is fundamental for solving a wide range of calculus problems.

2) Solid Cylinder Optimization

This section addresses an optimization problem involving a solid cylinder. Optimization problems are a cornerstone of calculus and its applications, particularly in fields like engineering, physics, and economics. These problems typically involve finding the maximum or minimum value of a function subject to certain constraints. In this case, we are dealing with a solid cylinder of a fixed volume, and we want to minimize its surface area. This is a classic optimization problem that demonstrates how calculus can be used to solve practical real-world problems. We will first express the surface area of the cylinder in terms of its radius, considering the constraint of the fixed volume. This will involve using the formula for the volume of a cylinder to eliminate one variable (the height) and express the surface area as a function of a single variable (the radius). Once we have the surface area as a function of the radius, we can use calculus techniques, such as finding critical points by setting the derivative equal to zero, to determine the radius that minimizes the surface area. This process will illustrate the power of calculus in finding optimal solutions to engineering design problems.

2.1) Expressing Surface Area ${A}$ in Terms of ${r}$

Given a solid cylinder with radius ${r}$ cm and height ${h}$ cm, the volume ${V}$ is given by the formula:

${V = \pi r^2 h}$

We are given that the volume ${V = 500}$ cm³. So,

${500 = \pi r^2 h}$

We can express the height ${h}$ in terms of the radius ${r}$ as:

${h = \frac{500}{\pi r^2}}$

The total surface area ${A}$ of the solid cylinder is given by the sum of the areas of the two circular ends and the lateral surface area:

${A = 2\pi r^2 + 2\pi rh}$

Now, we substitute the expression for ${h}$ in terms of ${r}$ into the surface area formula:

\begin{aligned} A &= 2\pi r^2 + 2\pi r \left(\frac{500}{\pi r^2}\right) \ &= 2\pi r^2 + \frac{1000}{r} \end{aligned}

Thus, the total surface area ${A}$ in terms of ${r}$ is:

${A(r) = 2\pi r^2 + \frac{1000}{r}}$

This equation expresses the surface area of the cylinder solely as a function of its radius, given the constraint of a fixed volume. This is a crucial step in optimization problems, as it allows us to use single-variable calculus techniques to find the minimum surface area. The ability to manipulate equations and express one variable in terms of another is a key skill in problem-solving, especially in mathematical contexts. By expressing the surface area as a function of the radius, we have set the stage for the next step, which involves finding the critical points of this function and determining the radius that minimizes the surface area.

2.2) Determining the Minimum Surface Area

To determine the value of ${r}$ that minimizes the surface area ${A(r) = 2\pi r^2 + \frac{1000}{r}}$, we need to find the critical points of the function. Critical points are the points where the derivative of the function is either zero or undefined. These points are potential locations of local minima or maxima. To find the critical points, we first need to find the derivative of ${A(r)}$ with respect to ${r}$:

\begin{aligned} \frac{dA}{dr} &= \frac{d}{dr} \left(2\pi r^2 + \frac{1000}{r}\right) \ &= 4\pi r - \frac{1000}{r^2} \end{aligned}

Now, we set the derivative equal to zero and solve for ${r}$:

${4\pi r - \frac{1000}{r^2} = 0}$

Multiplying through by ${r^2}$ gives:

${4\pi r^3 - 1000 = 0}$

Solving for ${r^3}$:

${4\pi r^3 = 1000}$

${r^3 = \frac{1000}{4\pi} = \frac{250}{\pi}}$

Taking the cube root:

${r = \sqrt[3]{\frac{250}{\pi}}}$

Now, we need to verify that this value of ${r}$ indeed minimizes the surface area. We can use the second derivative test for this. The second derivative of ${A(r)}$ is:

\begin{aligned} \frac{d^2A}{dr2} &= \frac{d}{dr} \left(4\pi r - \frac{1000}{r^2}\right) \ &= 4\pi + \frac{2000}{r^3} \end{aligned}

Since ${r}$ is a positive value, ${\frac{d^2A}{dr^2}}$ is always positive. This means that the function ${A(r)}$ is concave up, and the critical point we found corresponds to a minimum. Therefore, the radius that minimizes the surface area is:

${r = \sqrt[3]{\frac{250}{\pi}} \approx 4.301\text{ cm}}$

To find the corresponding height ${h}$, we substitute this value of ${r}$ back into the equation for ${h}$:

\begin{aligned} h &= \frac{500}{\pi r^2} \ &= \frac{500}{\pi \left(\sqrt[3]{\frac{250}{\pi}}\right)^2} \ &= \frac{500}{\pi \left(\frac{250}{\pi}\right)^{\frac{2}{3}}} \ &= \frac{500}{\pi \frac{250^{{\frac{2}{3}}}{\pi}{\frac{2}{3}}}} \ &= \frac{500 \pi^{\frac{2}{3}}}{\pi 250^{\frac{2}{3}}} \ &= \frac{500}{\pi^{\frac{1}{3}} 250^{\frac{2}{3}}} \ &= 2\sqrt[3]{\frac{250}{\pi}} \approx 8.602\text{ cm} \end{aligned}

Notice that ${h = 2r}$. This is a general result for cylinders: for a given volume, the surface area is minimized when the height is equal to the diameter. This result is significant because it provides a design principle for minimizing material usage in cylindrical structures. By understanding and applying this principle, engineers can optimize the dimensions of cylindrical containers and other structures, leading to cost savings and resource efficiency. The process of finding the minimum surface area involved using calculus techniques, including finding the first and second derivatives, setting the first derivative equal to zero, and using the second derivative test to confirm that we have found a minimum. These techniques are fundamental to optimization problems in calculus and have wide-ranging applications in various fields.

In summary, we have successfully found ${\frac{dy}{dx}}$ for the given functions using the product rule, quotient rule, and chain rule. We have also expressed the surface area of a solid cylinder in terms of its radius, given a fixed volume, and determined the dimensions that minimize the surface area. These examples highlight the power and versatility of calculus in solving both theoretical and practical problems. The techniques and concepts discussed in this article are essential for anyone studying calculus and its applications, and they provide a foundation for further exploration of more advanced topics in mathematics and related fields.