Question

Orthogonal Curves Two curves are said to cross at right angles if their tangents are perpendicular at the crossing point. The technical word for crossing at right angles is orthogonal. Show that the curves \(y=\sin 2 x\) and \(y=-\sin (x / 2)\) are orthogonal at the origin. Draw both graphs and both tangents in a square viewing window.

Short answer

The curves \(y = \sin 2x\) and \(y = -\sin(x/\2)\) are indeed orthogonal at the origin, as the product of their derivatives at that point is -1. Graph both curves and their tangents at the origin for visualization.

Step-by-step solution

Determine the first derivative of \(y =\sin 2x\)

The derivative of \(y = \sin 2x\) with respect to 'x', using the chain rule of differentiation, is \(y' = 2\cos 2x\)

Determine the first derivative of \(y = -\sin(x/2)\)

The derivative of \(y = -\sin(x/2)\), again using the chain rule of differentiation, is \(y' = -\frac{1}{2}\cos(x/2)\)

Evaluate the derivatives at the origin

Substitute x=0 in \(y' = 2\cos 2x\) and \(y' = -\frac{1}{2}\cos(x/2)\). The former gives \(2\cos(0) = 2\), while the latter gives \(-\frac{1}{2}\cos(0) = -\frac{1}{2}\)

Check for orthogonality

Two curves intersect orthogonally at a point if the product of their derivatives at that point is -1. Multiply the derivatives: \(2 \times -\frac{1}{2} = -1\), so they are orthogonal.

Draw the graphs

Graph the functions \(y = \sin 2x\) and \(y = -\sin(x/2)\) on the same plot. Their tangents at the origin are the lines given by their respective derivatives. A square viewing window with suitable dimensions should be used for a clear view.

Key concepts

Calculus

Calculus is an extensive area of mathematics that deals with the calculation of instantaneous rates of change (derivatives) and the summation of infinitely many small factors to determine a whole (integral). In the context of orthogonal curves, calculus is used to determine the gradient of the curves at any given point via derivatives. Orthogonal curves intersect at a right angle, and their tangents at the point of intersection must have slopes that are negative reciprocals of each other.

Specifically, when examining two curves for orthogonality, as in the exercise, we use differential calculus to find the slope of each curve at the intersecting point. If the product of these slopes is -1, then we can conclude the curves are orthogonal.

Chain Rule of Differentiation

The Chain Rule is a fundamental technique in calculus for finding the derivative of composite functions. When a function's variable is itself another function of a variable, such as in the given exercise with the functions involving trigonometric and multiplicative compositions, the Chain Rule comes into play. It specifies that the derivative of a compound function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

In our case, to differentiate y = sin(2x) and y = -sin(x/2), we apply the Chain Rule by finding the derivatives of the outer sine function and the inner functions 2x and x/2, respectively. This application of the Chain Rule in multiple steps helps unravel more complex derivatives.

Derivatives

Derivatives are a core concept of calculus and represent the rate at which a function is changing at any given point. In practical terms, the derivative of a function at any point gives the slope of the tangent line to the function's graph at that point. They are crucial for determining many properties of a function, including the motion along a curve, maxima and minima, and the behavior of the function's graph.

As part of the solution to determine if two curves are orthogonal, we calculate the first derivatives of the given functions. The first derivatives, evaluated at the point of intersection, reveal the slopes of the tangent lines. By examining whether these tangents are perpendicular, we can determine if the curves intersect orthogonally.

Trigonometric Functions

Trigonometric functions such as sine and cosine are periodic, and they appear frequently in the study of waves, oscillations, and circles. In calculus, these functions often describe the motion or shape of a curve. When we take derivatives of trigonometric functions, specific rules apply. The derivative of sine is cosine, and the derivative of cosine is the negative sine.

These functions are essential in the exercise because we use the derivatives of y = sin(2x) and y = -sin(x/2) to find the slope of their curves at the origin. By confirming that these slopes are negative reciprocals of each other (the product of their derivatives equals -1), we can conclude that the curves are orthogonal at the origin.