Question

Show that the curves \(y=\sqrt{2} \sin x\) and \(y=\sqrt{2} \cos x\) intersect at right angles at a certain point with \(0

Short answer

The curves intersect at a right angle at \(x=\frac{\pi}{4}\).

Step-by-step solution

Analyze Intersection Points

To find where the curves intersect, set their equations equal: \[ \sqrt{2} \sin x = \sqrt{2} \cos x \]Dividing both sides by \(\sqrt{2}\) gives:\[ \sin x = \cos x \]This implies that \( \tan x = 1 \), so \( x = \frac{\pi}{4} \).

Calculate Derivatives

Find the derivatives of both functions to evaluate their slopes at intersection. For \( y = \sqrt{2} \sin x \), the derivative is:\[ y' = \sqrt{2} \cos x \]For \( y = \sqrt{2} \cos x \), the derivative is:\[ y' = -\sqrt{2} \sin x \].

Evaluate Slopes at Intersection

Plug \( x = \frac{\pi}{4} \) into each derivative:- For \( y = \sqrt{2} \sin x \): \[ y' = \sqrt{2} \cdot \cos\left(\frac{\pi}{4}\right) = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \]- For \( y = \sqrt{2} \cos x \): \[ y' = -\sqrt{2} \cdot \sin\left(\frac{\pi}{4}\right) = -\sqrt{2} \cdot \frac{1}{\sqrt{2}} = -1 \].

Verify Intersection is Orthogonal

The product of slopes at the intersection point should be \(-1\) for curves to be orthogonal (right-angled). Check:\[ 1 \cdot (-1) = -1 \]The slope's product confirms the curves intersect at right angles.

Key concepts

Intersection Points

When two curves intersect, they share a common point on the graph. To discover this point, we need to set their equations equal to one another and solve for the variable. In our exercise, we're dealing with the curves given by:

• \( y = \sqrt{2} \sin x \)
• \( y = \sqrt{2} \cos x \)

Once we equate these functions and divide through by the common factor \(\sqrt{2}\), we discover that:\[ \sin x = \cos x \]This implies that the tangent of the angle \(x\) is equal to 1, which gives us the specific solution:\[ x = \frac{\pi}{4} \]The intersection point of these curves is crucial as it marks where these functions have the same output for a given input. This shared point will later be used to check their orthogonality.

Derivatives of Trigonometric Functions

Understanding derivatives is key in calculus, especially when finding the rate of change or the slope of functions. Trigonometric functions are a staple in calculus, and their derivatives follow specific rules. For our curves:

• The derivative of \( y = \sqrt{2} \sin x \) is the rate of change in reaction to changes in \(x\). By applying the derivative of \(\sin x\) which is \(\cos x\):\[ y' = \sqrt{2} \cos x \]
• Similarly, the derivative of \( y = \sqrt{2} \cos x \) involves the derivative of \(\cos x\), which is \(-\sin x\):\[ y' = -\sqrt{2} \sin x \]

Derivatives tell us about the slope of these curves, which is vital for determining angles at intersection points. Trigonometric derivatives like these need careful evaluation, especially at critical points such as \(x = \frac{\pi}{4}\).

Calculus Problem Solving

Solving calculus problems often requires step-by-step analysis and understanding of several calculus concepts such as differentiation and critical points. A structured approach can greatly aid in tackling complex exercises:

• **Identify equations and points of interest:** In this problem, we identified that we wanted to find where the curves \( y = \sqrt{2} \sin x \) and \( y = \sqrt{2} \cos x \) intersect.
• **Calculate derivatives:** These provide information about the slopes of the curves which are crucial for understanding how the curves intersect.
• **Evaluate at specific points:** Once the derivatives are determined, they are evaluated at the intersection points, in this case, \( x = \frac{\pi}{4} \), to assess relationships such as angles or relative slopes.

Breaking down each step simplifies what could otherwise be an overwhelming problem, allowing someone to proceed with clear, logical steps.

Slope of a Curve

The slope of a curve at any given point can be thought of as the 'steepness' or rate of change of the curve at that point. This is found using the derivative of the curve's equation. In the solution given:

• For \( y = \sqrt{2} \sin x \), the slope at the intersection point \( x = \frac{\pi}{4} \) is calculated as follows:\[ y' = \sqrt{2} \cdot \cos\left(\frac{\pi}{4}\right) = 1 \]
• For \( y = \sqrt{2} \cos x \), the slope at the same point is:\[ y' = -\sqrt{2} \cdot \sin\left(\frac{\pi}{4}\right) = -1 \]

When checking orthogonality (right angles) at the intersection point, the product of their slopes should be \(-1\). This confirms that the lines are perpendicular: \[ 1 \times (-1) = -1 \]The concept of slopes and derivatives is central to understanding not only intersection points but also the behavior of curves in calculus.