Problem 35 - 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. $$ \varphi_{1}(t)=t^{2}+1, \varphi_{2}(t)=t^{3}+1 \quad t_{0}=1 $$

Step-by-Step Solution

Answer

The tangent line is \( y = \frac{3}{2}x - 1 \).

Step 1: Find Derivatives of Functions

First, we need to find the derivatives of the functions: \( \varphi_1(t) = t^2 + 1 \) and \( \varphi_2(t) = t^3 + 1 \). Derivatives are:- \( \frac{dx}{dt} = \frac{d}{dt}(t^2 + 1) = 2t \)- \( \frac{dy}{dt} = \frac{d}{dt}(t^3 + 1) = 3t^2 \)

Step 2: Evaluate Derivatives at \(t_0\)

Next, we evaluate these derivatives at \( t_0 = 1 \).- \( \frac{dx}{dt} \bigg|_{t=1} = 2 \cdot 1 = 2 \)- \( \frac{dy}{dt} \bigg|_{t=1} = 3 \cdot 1^2 = 3 \)

Step 3: Compute Slope of Tangent Line

The slope \( m \) of the tangent line is computed using the derivative quotients: \[ m = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3}{2} \] So the slope of the tangent line at \( t_0 = 1 \) is \( \frac{3}{2} \).

Step 4: Find Point on Curve

To find the point where the tangent touches the curve, substitute \( t_0 = 1 \) into the parametric equations:- \( x = \varphi_1(1) = 1^2 + 1 = 2 \)- \( y = \varphi_2(1) = 1^3 + 1 = 2 \)Thus, the point is \( (2, 2) \).

Step 5: Write Equation of Tangent Line

Using the point-slope form of the line equation \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the point \((2, 2)\) and \(m = \frac{3}{2} \), the equation is:\[ y - 2 = \frac{3}{2}(x - 2) \] Simplifying gives \( y = \frac{3}{2}x - 1 \).

Key Concepts

Tangent Line

A tangent line to a curve is a straight line that touches the curve at a single point without crossing it. This line represents the direction in which the curve is heading at that point. In the context of parametric equations, like the ones given in the exercise, each part of the curve is described in terms of a parameter, usually denoted by \(t\). The tangent line is crucial in calculus because it gives us an idea of the instantaneous rate of change or how fast something is changing at any given point. For the problem at hand, the tangent line is calculated at the specific point \((2, 2)\) on the curve, which corresponds to the parameter value \(t_0 = 1\).

The tangent line for this solution is derived from the slope obtained through derivatives, highlighting the connection between differentiation and geometrical concepts of slopes and lines.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate of change of a function concerning its variable. When dealing with parametric equations, each component function has its derivative. In our exercise, these are the derivatives of the functions \(\varphi_1(t) = t^2 + 1\) and \(\varphi_2(t) = t^3 + 1\).

To find the derivatives, you compute:

• \( \frac{dx}{dt} = 2t \)
• \( \frac{dy}{dt} = 3t^2 \)

These expressions tell us how the \(x\) and \(y\) coordinates change as \(t\) changes. Evaluating these derivatives at \(t = 1\) gives us specific rates of change at that exact instance.

Such calculations are essential when determining the tangent line to a curve, as they provide the necessary slope information. Understanding derivatives helps in grasping the dynamic nature of equations and shapes.

Slope of Tangent

The slope of a tangent line is a measure of how steep the tangent is at the point where it touches the curve. It can be thought of as the ratio of the vertical change to the horizontal change at that point. In parametric equations, the slope \(m\) is found as the ratio of the derivatives:

• \(m = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

For our curve, substituting the derivatives evaluated at \(t = 1\) gives:

• \(m = \frac{3}{2}\)

This positive slope indicates that the tangent line rises as it moves from left to right.

The slope of the tangent is not only pivotal for determining the tangent line equation but also provides insights into the behavior of the function at specific points, such as whether it's increasing or decreasing.

Point-Slope Form

The point-slope form of a linear equation is a convenient way to write the equation of a line when you know both a point on the line and the slope. It is expressed as:

• \(y - y_1 = m(x - x_1)\)

Here, \((x_1, y_1)\) is any specific point on the line, and \(m\) is the slope. For our example, the point \((2, 2)\) is used as \((x_1, y_1)\) with a slope of \(\frac{3}{2}\). Therefore, the point-slope form of the tangent line at the curve is:

• \( y - 2 = \frac{3}{2}(x - 2) \)

This form makes it simple to substitute known values and solve for the equation of the tangent line.

This method of equation writing is particularly useful in calculus and geometry for swiftly deriving the equation of lines with given parameters.