Problem 82 - Calculus Volume 3

Question

Find the equation of the tangent line to \(x=\sin (t), y=\cos (t)\) at \(t=\frac{\pi}{4}\).

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = -x + \sqrt{2} \).

Step 1: Identify the Parametric Equations

The problem gives us the parametric equations: \( x = \sin(t) \) and \( y = \cos(t) \). We need to find the equation of the tangent line at \( t = \frac{\pi}{4} \).

Step 2: Calculate Derivatives of the Parametric Equations

To find the slope of the tangent, we need the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).- \( \frac{dx}{dt} = \cos(t) \).- \( \frac{dy}{dt} = -\sin(t) \).

Step 3: Calculate Slope of Tangent Line

The slope \( m \) of the tangent line to the curve can be found by \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \).Substituting the derivatives, we get:\[ \frac{dy}{dx} = \frac{-\sin(t)}{\cos(t)} = -\tan(t). \]At \( t = \frac{\pi}{4} \), \( \tan \left(\frac{\pi}{4}\right) = 1 \), hence, \( \frac{dy}{dx} = -1 \).

Step 4: Calculate Point on the Curve

We need the coordinates of the point on the curve at \( t = \frac{\pi}{4} \):- \( x = \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \).- \( y = \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \).Thus, the point is \( \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \).

Step 5: Write the Equation of the Tangent Line

Using the point-slope form of a line \( y - y_1 = m(x - x_1) \), where \( m = -1 \) and \( (x_1, y_1) = \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \):\[ y - \frac{\sqrt{2}}{2} = -1 \left( x - \frac{\sqrt{2}}{2} \right) \].Simplifying gives:\[ y = -x + \sqrt{2} \].

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at just one specific point. This means it matches the direction of the curve at that point. Think of a tangent as a straight line that gently "kisses" the curve exactly where they meet, without crossing it. Finding the equation of a tangent line is crucial to understand how a curve behaves at a specific point.
This is particularly useful in various applications, such as physics and engineering, where the slope of the tangent line tells you how a specific quantity is changing at that precise point.To find the tangent line of a curve defined by parametric equations like in our example, where

• x = \( \sin(t) \)
• y = \( \cos(t) \)

one needs to calculate the slope of this tangent. The slope will help derive the tangent line’s equation that describes how the line interacts with the curve at a given point. Here, at \( t = \frac{\pi}{4} \), the tangent touches the curve, offering insight into the curve's instantaneous direction.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which things change. In simpler terms, a derivative can indicate how one quantity changes with respect to another. For curves described using parametric equations, we have derivatives with respect to the parameter, in this case, 't'.
Calculating derivatives of the parametric equations helps us understand the rate of change along the x-axis and y-axis. Differentiate the parametric equations:

• \( \frac{dx}{dt} = \cos(t) \)
• \( \frac{dy}{dt} = -\sin(t) \)

Knowing these derivatives, we can establish how one variable changes relative to the other. After finding these individual rates of change, we combine them to find the derivative \( \frac{dy}{dx} \), which indicates how 'y' changes with 'x'. This derived slope \(-\tan(t)\) is what allows us to define the behavior of the tangent line. So, when \( t = \frac{\pi}{4} \), we substitute this into our derivative \( \frac{dy}{dx} \) to find that the slope is -1.

Point-Slope Form

One of the easiest ways to write the equation of a line is using the point-slope form. This form is highly applicable when you know:

• A point on the line \((x_1, y_1)\)
• The slope of the line 'm'

The formula looks like this: \[ y - y_1 = m(x - x_1) \] Let's apply this to our problem. We know the point on the curve at \( t = \frac{\pi}{4} \) is \( \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \), and our slope \( m = -1 \). Plug these values into the formula:\[ y - \frac{\sqrt{2}}{2} = -1 \times (x - \frac{\sqrt{2}}{2}) \]Rearranging this, we find the equation of our tangent line is \( y = -x + \sqrt{2} \). This equation gives us a precise description of the tangent line, showing exactly how it's positioned in relation to the curve at that point.