Problem 26 - Advanced Engineering Mathematics

Question

In Problems, find parametric equations of the tangent line to the given curve at the indicated value of \(t\). $$ x=t^{3}-t, y=\frac{6 t}{t+1}, z=(2 t+1)^{2} ; t=1 $$

Step-by-Step Solution

Answer

Parametric equations: \(x = 2t\), \(y = 3 + \frac{3}{2}t\), \(z = 9 + 12t\).

Step 1: Determine the point on the curve at t = 1

To find the point on the curve when \( t = 1 \), substitute \( t = 1 \) into the parametric equations. \[ x = 1^3 - 1 = 0 \]\[ y = \frac{6 \times 1}{1 + 1} = 3 \]\[ z = (2 \times 1 + 1)^2 = 9 \]Therefore, the point on the curve is \((0, 3, 9)\).

Step 2: Find the derivative with respect to t for each coordinate

Calculate the derivatives to determine the direction vector of the tangent line:\[ \frac{dx}{dt} = 3t^2 - 1 \]\[ \frac{dy}{dt} = \frac{6(1)}{(t+1)^2} = \frac{6}{(t+1)^2} \] (using the quotient rule)\[ \frac{dz}{dt} = 2(2t+1)(2) = 8t+4 \] These derivatives represent the velocities in the respective coordinates.

Step 3: Evaluate the derivatives at t = 1

Substitute \( t = 1 \) into the derivatives to find the tangent vector at this point:\[ \frac{dx}{dt} = 3(1)^2 - 1 = 2 \]\[ \frac{dy}{dt} = \frac{6}{(1+1)^2} = \frac{3}{2} \]\[ \frac{dz}{dt} = 8(1) + 4 = 12 \]The tangent vector at \( t = 1 \) is \((2, \frac{3}{2}, 12)\).

Step 4: Write the parametric equations for the tangent line

The parametric equations for the tangent line passing through the point \((0, 3, 9)\) with direction vector \((2, \frac{3}{2}, 12)\) are:\[ x(t) = 0 + 2t \]\[ y(t) = 3 + \frac{3}{2}t \]\[ z(t) = 9 + 12t \]These are the parametric equations for the tangent line.

Key Concepts

Parametric Equations

In mathematics, parametric equations represent a set of related quantities as functions of a common variable. Often, this variable is denoted by \( t \), commonly referred to as the parameter. The role of parametric equations is crucial when dealing with curves in space because they enable us to define each coordinate — \( x \), \( y \), and \( z \) — as separate functions of \( t \). This is particularly useful when a curve cannot be easily described using a single equation in terms of \( x \) and \( y \) only.

For example, in the problem provided, the given parametric equations \( x = t^3 - t \), \( y = \frac{6t}{t+1} \), and \( z = (2t+1)^2 \) express each of the coordinates of a point on the curve in terms of the parameter \( t \). These allow you to find precise points on the curve for specific values of \( t \). When solving problems involving parametric equations, it is key to understand how to manipulate these parameters to extract meaningful results, such as the position on the curve at a particular time or location.

Derivative

Derivatives are fundamental in calculus, helping us understand the rate of change of functions. When dealing with parametric equations, we derive each function concerning \( t \). This approach gives us the components of the velocity vector that outlines how the point on the curve changes as \( t \) changes.

In the exercise, derivatives \( \frac{dx}{dt} = 3t^2 - 1 \), \( \frac{dy}{dt} = \frac{6}{(t+1)^2} \), and \( \frac{dz}{dt} = 8t + 4 \) were calculated for the parametric equations. These denote the rates of change for each of the \( x \), \( y \), and \( z \) coordinates. Evaluating these derivatives at \( t = 1 \) gives \( 2 \), \( \frac{3}{2} \), and \( 12 \) respectively, forming the direction vector. The direction vector is crucial in establishing the tangent line.

Direction Vector

When finding the tangent line to a curve at a specific point, the direction vector becomes important. This vector indicates the direction in which the line tangent to the curve is pointing. It’s derived from the derivatives of the parametric equations evaluated at the specific parameter \( t \).

As per the step-by-step solution, the derivatives at \( t = 1 \) were found to be \( (2, \frac{3}{2}, 12) \), which serves as the direction vector for the tangent line. To express the tangent line using parametric equations, this vector is combined with the point on the curve — \( (0, 3, 9) \). The resulting tangent line can be represented as:

• \( x(t) = 0 + 2t \)
• \( y(t) = 3 + \frac{3}{2}t \)
• \( z(t) = 9 + 12t \)

Understanding the direction vector and its application allows you to succinctly express the linear approximation of a curve at a given point, crucial in fields such as physics and engineering.