Problem 38 - Calculus Single Variable

Question

Find the tangent line to the parametric curve \(x=\varphi_{1}(t), y=\varphi_{2}(t)\) at the point corresponding to the given value \(t_{0}\) of the parameter. $$ \varphi_{1}(t)=\cos (2 t), \varphi_{2}(t)=\sin (3 t) \quad t_{0}=\pi / 4 $$

Step-by-Step Solution

Answer

The tangent line is \(y = \frac{3\sqrt{2}}{4}x + \frac{\sqrt{2}}{2}\).

Step 1: Find Derivatives of the Parametric Equations

To find the equation of the tangent line, we first need to determine the derivatives of \(x\) and \(y\) with respect to \(t\). Differentiating \(x = \varphi_1(t) = \cos(2t)\), we have \(\frac{dx}{dt} = -2\sin(2t)\). Similarly, differentiating \(y = \varphi_2(t) = \sin(3t)\), we obtain \(\frac{dy}{dt} = 3\cos(3t)\).

Step 2: Evaluate Derivatives at t_0

We now evaluate these derivatives at the given \(t_0 = \frac{\pi}{4}\). For \(\frac{dx}{dt}\), substitute \(t = \frac{\pi}{4}\): \(-2\sin(\frac{\pi}{2}) = -2\). For \(\frac{dy}{dt}\), substitute \(t = \frac{\pi}{4}\): \(3\cos(\frac{3\pi}{4}) = 3\left(-\frac{\sqrt{2}}{2}\right) = -\frac{3\sqrt{2}}{2}\).

Step 3: Find the Slope of the Tangent Line

The slope of the tangent line to the curve at \(t_0\) is given by \(m = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\). Substituting the derivatives, we have \(m = \frac{-\frac{3\sqrt{2}}{2}}{-2} = \frac{3\sqrt{2}}{4}\).

Step 4: Evaluate the Parametric Equations at t_0

Next, we find the coordinates of the point on the curve at \(t_0 = \frac{\pi}{4}\). For \(x\), substitute \(t = \frac{\pi}{4}\): \(\cos(\frac{\pi}{2}) = 0\). For \(y\), substitute \(t = \frac{\pi}{4}\): \(\sin(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2}\). Therefore, the point is \((0, \frac{\sqrt{2}}{2})\).

Step 5: Write the Equation of the Tangent Line

The equation of the tangent line can be written in point-slope form: \(y - y_1 = m(x - x_1)\). Here, \((x_1, y_1) = (0, \frac{\sqrt{2}}{2})\) and \(m = \frac{3\sqrt{2}}{4}\). Substituting these values, we get \(y - \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{4}(x - 0)\), which simplifies to \(y = \frac{3\sqrt{2}}{4}x + \frac{\sqrt{2}}{2}\).

Key Concepts

Parametric Equations

Parametric equations are a way of expressing a curve by defining both the x and y coordinates as functions of a third variable, often denoted as t. This method is particularly useful for describing curves that do not represent functions in the traditional sense.
In our exercise, we have the parametric equations given as \( x = \varphi_{1}(t) = \cos(2t) \) and \( y = \varphi_{2}(t) = \sin(3t) \). Here, the variable \( t \) is called the parameter, and for each value of \( t \), there is a corresponding point on the curve.

• The advantage of parametric equations is their capability to describe complex curves like ellipses, circles, and others which are not functions when using the traditional \( y=f(x) \) form.
• They provide great flexibility in modeling dynamic systems where time or another variable is involved as a parameter.

Finding Derivatives

To determine the characteristics of a curve described by parametric equations, such as its tangent line, we need to calculate derivatives with respect to the parameter \( t \). These derivatives help us understand how the x and y coordinates change as \( t \) changes.
For the given parametric equations:

• The derivative of \( x \) with respect to \( t \) is \( \frac{dx}{dt} = -2\sin(2t) \). This represents the change in the x-coordinate as \( t \) varies.
• Similarly, the derivative of \( y \) with respect to \( t \) is \( \frac{dy}{dt} = 3\cos(3t) \), indicating the rate of change of the y-coordinate with \( t \).

Understanding these derivatives is pivotal in determining the properties of the curve, including the slope of the tangent line at a specific point.

Slope of Tangent Line

The slope of a tangent line to a curve at a given point is a critical concept in calculus. It is found using the derivatives of the parametric equations. The formula for calculating the slope \( m \) of a tangent line at a specific \( t \) is:\[m = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{3\cos(3t)}{-2\sin(2t)}\]In our exercise, after evaluating at \( t_0 = \frac{\pi}{4} \):

• \( \frac{dx}{dt} = -2 \)
• \( \frac{dy}{dt} = -\frac{3\sqrt{2}}{2} \)
• Thus, \( m = \frac{\frac{-3\sqrt{2}}{2}}{-2} = \frac{3\sqrt{2}}{4} \)

The slope tells us how steep the tangent line is and the direction in which it slants. A positive slope indicates an upward slant, while a negative slope means it slants downwards.

Point-Slope Form of a Line

The point-slope form is a useful algebraic method for expressing the equation of a line that passes through a given point with a known slope. This form is particularly convenient for writing equations of tangent lines to curves.
The general equation is:\[y - y_1 = m(x - x_1)\]Where:

• \( m \) is the slope of the line.
• \( (x_1, y_1) \) is the specific point on the line.

Given the point \( (0, \frac{\sqrt{2}}{2}) \) and the slope \( \frac{3\sqrt{2}}{4} \), we can plug these into the formula to find the equation of the tangent line:

• \( y - \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{4}(x - 0) \)

This simplifies to:\[y = \frac{3\sqrt{2}}{4}x + \frac{\sqrt{2}}{2}\]Using this form, you can easily see how the line behaves relative to the curve at the given point.