Problem 27 - Thomas Calculus

Question

In Exercises \(27-30\) , find the value(s) of \(t\) so that the tangent line to the given curve contains the given point. $$ \mathbf{r}(t)=t^{2} \mathbf{i}+(1+t) \mathbf{j}+(2 t-3) \mathbf{k} ; \quad(-8,2,-1) $$

Step-by-Step Solution

Answer

The values of \(t\) are 4 and -2.

Step 1: Determine the Parametric Equations

First, let's express the vector function \(\mathbf{r}(t) = t^2 \mathbf{i} + (1+t) \mathbf{j} + (2t-3) \mathbf{k}\) in terms of its parametric form. The parametric equations for the coordinates are given as follows: \(x(t) = t^2\), \(y(t) = 1 + t\), and \(z(t) = 2t - 3\).

Step 2: Find the Tangent Vector

To find the tangent vector at any point \(t\), we need to take the derivative of \(\mathbf{r}(t)\) with respect to \(t\). This gives us \(\mathbf{r}'(t) = \frac{d}{dt}(t^2) \mathbf{i} + \frac{d}{dt}(1+t) \mathbf{j} + \frac{d}{dt}(2t-3) \mathbf{k} = 2t \mathbf{i} + 1 \mathbf{j} + 2 \mathbf{k}\).

Step 3: Write the Equation of the Tangent Line

Using the point \(P = \mathbf{r}(t)\) on the curve, the equation of the tangent line can be written in vector form as \(\mathbf{L}(t) = \mathbf{r}(t) + s \mathbf{r}'(t)\), where \(s\) is a scalar parameter. This becomes \((t^2 + 2ts)\mathbf{i} + (1+t+s)\mathbf{j} + ((2t-3)+2s)\mathbf{k}\).

Step 4: Equate to Given Point

We set the coordinates of \(\mathbf{L}(t)\) equal to the given point \((-8,2,-1)\). This gives us the system of equations: \(t^2 + 2ts = -8\), \(1 + t + s = 2\), and \((2t - 3) + 2s = -1\).

Step 5: Solve the System of Equations

Solve the system of equations: 1) \(t^2 + 2ts = -8\), 2) \(1 + t + s = 2\), 3) \(2t - 3 + 2s = -1\). First simplify equation (2) to get \(s = 1 - t\). Substitute \(s\) into the other equations, yielding: \(t^2 + 2t(1-t) = -8\) and \(2t - 3 + 2(1-t) = -1\). Simplify further to solve for \(t\).

Step 6: Final Simplifications and Solutions

Substitute \(s = 1-t\) into the equations, simplifying equation (3) gives \(2t - 3 + 2 - 2t = -1\), which holds true for any \(t\). From \(t^2 + 2t - 2t^2 = -8\), simplify to \(-t^2 + 2t + 8 = 0\) becoming \(t^2 - 2t - 8 = 0\). Solve the quadratic equation to find \(t = 4\) or \(t = -2\).

Step 7: Verify the Solution

Verify by plugging \(t=4\) and \(t=-2\) back into the original parametric equations to ensure it satisfies the conditions. For \(t = 4\), \(s = -3\), transforms each equation correctly. For \(t = -2\), \(s = 3\), also transforms each equation correctly. Both values satisfy the conditions.

Key Concepts

Parametric Equations

Parametric equations are a set of equations that describe a curve or surface by expressing the coordinates of the points on the curve as functions of one or more parameters. In this exercise, the vector function is provided as \( \mathbf{r}(t) = t^2 \mathbf{i} + (1+t) \mathbf{j} + (2t-3) \mathbf{k} \). This vector function can be broken down into three parametric equations that specify each dimension:

• \(x(t) = t^2\) for the x-coordinate,
• \(y(t) = 1 + t\) for the y-coordinate,
• \(z(t) = 2t - 3\) for the z-coordinate.

To understand the behavior of the curve, we vary the parameter \(t\), which then determines the point \((x(t), y(t), z(t))\) on the curve. This allows for a comprehensive representation of complex shapes in a more manageable and calculable form.

Tangent Vector

A tangent vector is critical in understanding how a curve changes at a particular point. It gives the direction in which the curve progresses and is found by taking the derivative of the vector function with respect to the parameter \(t\). In the context of this exercise, the tangent vector \( \mathbf{r}'(t) \) is derived by differentiating \( \mathbf{r}(t) \):

• The derivative of the x-component \( t^2 \) is \( 2t \).
• The derivative of the y-component \( 1 + t \) is \( 1 \).
• The derivative of the z-component \( 2t - 3 \) is \( 2 \).

Thus, the tangent vector is \( \mathbf{r}'(t) = 2t \mathbf{i} + 1 \mathbf{j} + 2 \mathbf{k} \). This vector not only points in the direction of the curve at point \(t\) but also provides the slope information needed to write the equation of the tangent line.

System of Equations

Solving a system of equations is an essential skill for progressing through this problem. A system of equations consists of multiple equations that are all interconnected through shared variables. Here, you have three equations:

• \(t^2 + 2ts = -8\),
• \(1 + t + s = 2\),
• \(2t - 3 + 2s = -1\).

The goal is to determine values for \(t\), the parameter, and \(s\), the scalar, that satisfy all equations simultaneously. This involves a clever substitution of \(s = 1 - t\) (from the simplified second equation) into the others, helping to isolate \(t\) and solve the system efficiently. By solving this system, we determine the correct values of \(t\) where the tangent line passes through the given point \((-8, 2, -1)\).

Quadratic Equation

A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\), which typically appears when solving problems involving parabolas or curves. In this exercise's context, we arrive at a quadratic equation after substitutions that involve \(t\): \(t^2 - 2t - 8 = 0\). To solve this quadratic equation, one can use various methods:

• The quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\),
• Factoring, if the equation is easily simplified,
• Or completing the square.

In this instance, by solving \(t^2 - 2t - 8 = 0\), we find the solutions \(t = 4\) and \(t = -2\). Both values satisfy the conditions of the original problem after verification, showing that the tangent line intersects the given point for these \(t\) values.