Problem 4 - Active Calculus Multivariable

Question

Find the curvature of \(y=\sin (-2 x)\) at \(x=\frac{\pi}{4}\).

Step-by-Step Solution

Answer

The curvature of the function \(y = \sin(-2x)\) at \(x = \frac{\pi}{4}\) is \(K\left(\frac{\pi}{4}\right) = \frac{8\sqrt{125}}{125}\).

Step 1: Find the first derivative of the function.

Differentiate the function \(y = \sin(-2x)\) with respect to \(x\) using the chain rule: \[y'(x) = \frac{dy}{dx} = \frac{d(\sin(-2x))}{dx} = \cos(-2x)(-2)\] So we have: \[y'(x) = -2 \cos (-2x)\]

Step 2: Find the second derivative of the function.

Let's differentiate the first derivative we found in the previous step: \[y''(x) = \frac{d(-2\cos(-2x))}{dx} = 4\sin(-2x)(-2)\] So we have: \[y''(x) = -8\sin(-2x)\]

Step 3: Calculate the first and second derivatives at \(x = \frac{\pi}{4}\).

Substitute the value \(x = \frac{\pi}{4}\) into the first and second derivatives to find their values at this point: \[y'\left(\frac{\pi}{4}\right) = -2 \cos \left(-2\cdot \frac{\pi}{4}\right) = -2 \cos (-\frac{\pi}{2}) = -2\] Similarly, evaluate the second derivative: \[y''\left(\frac{\pi}{4}\right) = -8\sin\left(-2\cdot\frac{\pi}{4}\right) = -8\sin(-\frac{\pi}{2}) = 8\]

Step 4: Use the curvature formula.

We'll plug the values we found into the curvature formula: \[K(x) = \frac{|y''(x)|}{(1+(y'(x))^2)^{3/2}}\] Substitute \(y'(\frac{\pi}{4}) = -2\) and \(y''(\frac{\pi}{4}) = 8\) into the formula: \[K\left(\frac{\pi}{4}\right) = \frac{|8|}{(1+(-2)^2)^{3/2}} = \frac{8}{(1+4)^{3/2}}\]

Step 5: Simplify the curvature at \(x = \frac{\pi}{4}\).

Simplify the expression for the curvature at \(x = \frac{\pi}{4}\): \[K\left(\frac{\pi}{4}\right) = \frac{8}{\sqrt{5^3}} = \frac{8}{\sqrt{125}} = \frac{8\sqrt{125}}{125}\] So the curvature of function \(y = \sin(-2x)\) at \(x = \frac{\pi}{4}\) is: \[K\left(\frac{\pi}{4}\right) = \frac{8\sqrt{125}}{125}\]

Key Concepts

Differentiation

Differentiation is a core mathematical concept used to determine the rate at which a function is changing at any given point. In simpler terms, it is the process of finding the derivative of a function. A derivative represents an instantaneous rate of change, much like how speed indicates how fast an object moves. For the function \( y = \sin(-2x) \), the task involves finding the first and second derivatives necessary to determine the curvature.To differentiate \( y = \sin(-2x) \), we used basic differentiation rules, combined with the chain rule. The first derivative, \( y'(x) = -2\cos(-2x) \), tells us about the slope of the tangent to the curve at any point \( x \). Keep in mind that a high level of accuracy in differentiation provides a more precise understanding of how a function behaves at any point along its curve.

Chain Rule

The chain rule is a vital differentiation technique used when dealing with composite functions, that is, functions within functions. It allows us to differentiate the outer function while considering the inner function.For the given function \( y = \sin(-2x) \), the chain rule comes into play because \( -2x \) is an inner function. When differentiating such an expression, you:

• Differentiate the outside function (\( \sin \) becomes \( \cos \))
• Multiply by the derivative of the inside function (the derivative of \( -2x \) is \( -2 \))

Thus, applying the chain rule, we arrive at the first derivative \( y'(x) = -2\cos(-2x) \), demonstrating the chain rule's efficiency in handling such nested functions. Remember, mastering the chain rule is essential for tackling more complex calculus problems.

Curvature Formula

Curvature provides a measure of how sharply a curve bends at a given point. For a function \( y = f(x) \), the formula for curvature, \( K(x) \), is given by:\[K(x) = \frac{|y''(x)|}{(1+(y'(x))^2)^{3/2}}\]This mathematical expression utilizes both the first and second derivatives of the function.To find the curvature of \( y = \sin(-2x) \) at \( x = \frac{\pi}{4} \), we substituted the values of the derivatives we previously found: \( y'\left(\frac{\pi}{4}\right) = -2 \) and \( y''\left(\frac{\pi}{4}\right) = 8 \). By plugging these into the curvature formula, you can calculate the curve's extent of bending at that specific point, resulting in \( K\left(\frac{\pi}{4}\right) = \frac{8\sqrt{125}}{125} \). Understanding curvature is essential for fields such as physics and engineering, where modeling natural phenomena require precision.

Trigonometric Functions

Trigonometric functions, such as \( \sin(x) \), \( \cos(x) \), and \( \tan(x) \), play vital roles in various mathematical applications. They describe relationships between the angles and sides of triangles, but they also extend to describe periodic phenomena such as waves.When differentiating trigonometric functions, it is important to note their specific derivative rules:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \(-\sin(x) \).
• The derivative of \( \tan(x) \) is \( \sec^2(x) \).

In our exercise, \( y = \sin(-2x) \), we observed trigonometric differentiation in action. The presence of the minus sign in \( -2x \) and subsequent application of the chain rule illustrates how trigonometric functions are often manipulated in calculus. Mastery of these functions and their derivatives is crucial for comprehensively tackling calculus problems.