Question

Suppose the vector-valued function \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) is smooth on an interval containing the point \(t_{0}\) The line tangent to \(\mathbf{r}(t)\) at \(t=t_{0}\) is the line parallel to the tangent vector \(\mathbf{r}^{\prime}\left(t_{0}\right)\) that passes through \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right) .\) For each of the following functions, find an equation of the line tangent to the curve at \(t=t_{0} .\) Choose an orientation for the line that is the same as the direction of \(\mathbf{r}^{\prime}.\) $$\mathbf{r}(t)=\left\langle 3 t-1,7 t+2, t^{2}\right\rangle ; t_{0}=1$$

Short answer

Question: Find an equation for the line tangent to the curve given by the vector-valued function \(\mathbf{r}(t) = \left\langle 3t - 1, 7t + 2, t^{2}\right\rangle\) at \(t = 1\). Answer: The equation of the tangent line is \(\mathbf{L}(t) = \left\langle 2 + 3t, 9 + 7t, 1 + 2t \right\rangle\).

Step-by-step solution

Compute the tangent vector \(\mathbf{r}'(t_0)\)

First, we need to compute the derivative of the given vector-valued function. Given that \(\mathbf{r}(t) = \left\langle 3t - 1, 7t + 2, t^{2}\right\rangle\), let's find its derivative \(\mathbf{r}'(t)\): $$\mathbf{r}'(t) = \frac{d}{dt} \langle 3t-1, 7t+2, t^2 \rangle = \langle \frac{d}{dt}(3t-1), \frac{d}{dt}(7t+2), \frac{d}{dt}(t^2) \rangle$$ Now, take the derivative of each component: $$\mathbf{r}'(t) = \langle 3, 7, 2t \rangle$$ Next, we need to find the tangent vector \(\mathbf{r}'(t_0)\). Given that \(t_0 = 1\), we get: $$\mathbf{r}'(t_0) = \langle 3, 7, 2(1) \rangle = \langle 3, 7, 2 \rangle$$

Calculate the point \(P\left(f(t_0), g(t_0), h(t_0)\right)\) on the curve

Now we need to find the point \(P\) on the curve when \(t = t_0 = 1\). Using the given function, we can plug in the value of \(t_0\) to obtain the coordinates of \(P\) : $$P\left(f(t_0), g(t_0), h(t_0)\right) = \left(f(1), g(1), h(1)\right) = \left(3(1) - 1, 7(1) + 2, (1)^2\right) = (2, 9, 1)$$

Write the equation for the tangent line using the tangent vector and the point

Finally, we can write the equation for the tangent line using the tangent vector from Step 1 and the point from Step 2. We use the point-vector form for the equation of a line: $$\mathbf{L}(t) = P + t\mathbf{r}'(t_0) = \langle 2, 9, 1 \rangle + t \langle 3, 7, 2 \rangle$$ Here is the equation of the tangent line: $$\mathbf{L}(t) = \langle 2 + 3t, 9 + 7t, 1 + 2t \rangle$$

Key concepts

Vector-Valued Functions

Vector-valued functions extend the idea of a function by assigning a vector, rather than a scalar, to every point in their domain. Imagine a traditional function that gives you a single number for every input. A vector-valued function, however, gives you a whole set of numbers—a vector—for every input. This is written as a vector \(\mathbf{r}(t)\) with components that are functions of the variable \(t\), like \(\langle f(t), g(t), h(t)\rangle\) in our example.

This concept is crucial when dealing with curves in space. The position of a point in three-dimensional space at any given time can be described by such a function. It's like describing the path of a spaceship: at each moment in time \(t\), the ship's coordinates are represented by a three-component vector. In the exercise, the function \(\mathbf{r}(t) = \left\langle 3t - 1, 7t + 2, t^{2}\right\rangle\) describes such a path in three dimensions.

Derivatives of Vector Functions

When we derive a scalar function, we're finding its rate of change at any point. This concept is similar for vector functions, but now we're interested in the rate of change for each component function inside the vector function. In essence, computing the derivative of a vector function involves differentiating each of the component functions individually.

For the vector function \(\mathbf{r}(t)\) from our exercise, the derivative is found by taking the derivative of each component function as though they were separate: \(\mathbf{r}'(t) = \langle 3, 7, 2t \rangle\). This derivative represents the tangent vector, which tells us the direction of the path at any instant. If this path represents our spaceship's journey, the tangent vector \(\mathbf{r}'(t)\) shows the direction it's pointing or moving at the moment \(t\).

The specific value \(\mathbf{r}'(t_0)\) when \(t = t_0\) is like taking a snapshot of the spaceship's direction at that particular moment, which is crucial for understanding the path's behavior at \(t_0\).

Equation of a Tangent Line

The equation of the tangent line to a curve at a given point is a linear approximation of the curve near that point. Think of it as zooming in so closely on a curve that it starts looking like a straight line. This line only touches the curve exactly at one point, which is why it's called 'tangent'.

To determine the equation of the tangent line, you need two things: a point on the curve and the tangent vector at that point. From our exercise, we have the point \( P = (2, 9, 1) \) and the tangent vector \(\mathbf{r}'(t_0) = \langle 3, 7, 2 \rangle\). Then, using the point-vector form, denoted \( \mathbf{L}(t) = P + t\mathbf{r}'(t_0)\), we find the equation for the tangent line.

The equation \(\mathbf{L}(t) = \langle 2 + 3t, 9 + 7t, 1 + 2t \rangle\) effectively tells us that for any small movement away from \(t_0\), this is where the spaceship will be, assuming it keeps moving in the same direction with no turns—its path traced out by the line.