Find all times t so that the tangent line to the parametric curve ⟨t − 8, t2 − 5t − 8, 48 − t2⟩ at
time t passes through the origin

The parametric curve is given by:

\[
\mathbf{r}(t) = \langle t - 8, t^2 - 5t - 8, 48 - t^2 \rangle
\]

We are tasked with finding all times \( t \) such that the tangent line to this curve at time \( t \) passes through the origin \((0, 0, 0)\).

### Step 1: Find the tangent vector

First, calculate the derivative of the parametric curve to find the tangent vector \( \mathbf{r}'(t) \):

\[
\mathbf{r}'(t) = \frac{d}{dt} \langle t - 8, t^2 - 5t - 8, 48 - t^2 \rangle
= \langle 1, 2t - 5, -2t \rangle
\]

So, the tangent vector at any time \( t \) is:

\[
\mathbf{r}'(t) = \langle 1, 2t - 5, -2t \rangle
\]

### Step 2: Equation of the tangent line

The parametric equation of the tangent line at time \( t \) is:

\[
\mathbf{r}(s) = \mathbf{r}(t) + s \mathbf{r}'(t)
\]

This represents a line passing through the point \( \mathbf{r}(t) = \langle t - 8, t^2 - 5t - 8, 48 - t^2 \rangle \) with direction given by \( \mathbf{r}'(t) = \langle 1, 2t - 5, -2t \rangle \).

Thus, the tangent line can be written as:

\[
\mathbf{r}(s) = \langle t - 8, t^2 - 5t - 8, 48 - t^2 \rangle + s \langle 1, 2t - 5, -2t \rangle
\]

The parametric form of the line is:

\[
\mathbf{r}(s) = \langle (t - 8) + s, (t^2 - 5t - 8) + s(2t - 5), (48 - t^2) - 2st \rangle
\]

### Step 3: Set the line to pass through the origin

For the tangent line to pass through the origin \((0, 0, 0)\), there must be some value of \( s \) such that:

\[
\mathbf{r}(s) = (0, 0, 0)
\]

This gives us the system of equations:

1. \((t - 8) + s = 0\)
2. \((t^2 - 5t - 8) + s(2t - 5) = 0\)
3. \((48 - t^2) - 2st = 0\)

### Step 4: Solve the system of equations

#### Equation 1:
From the first equation:

\[
(t - 8) + s = 0 \quad \Rightarrow \quad s = 8 - t
\]

#### Equation 2:
Substitute \( s = 8 - t \) into the second equation:

\[
(t^2 - 5t - 8) + (8 - t)(2t - 5) = 0
\]

Expand and simplify:

\[
t^2 - 5t - 8 + (8 - t)(2t - 5) = 0
\]
\[
t^2 - 5t - 8 + (16t - 40 - 2t^2 + 5t) = 0
\]
\[
t^2 - 5t - 8 + 16t - 40 - 2t^2 + 5t = 0
\]
\[
-t^2 + 16t - 48 = 0
\]
\[
t^2 - 16t + 48 = 0
\]

Solve this quadratic equation:

\[
t = \frac{16 \pm \sqrt{16^2 - 4(1)(48)}}{2(1)} = \frac{16 \pm \sqrt{256 - 192}}{2} = \frac{16 \pm \sqrt{64}}{2} = \frac{16 \pm 8}{2}
\]

Thus, the solutions for \( t \) are:

\[
t = \frac{16 + 8}{2} = 12 \quad \text{or} \quad t = \frac{16 - 8}{2} = 4
\]

#### Equation 3:
Finally, check if these values of \( t \) satisfy the third equation. Substituting \( s = 8 - t \) into the third equation:

\[
(48 - t^2) - 2t(8 - t) = 0
\]
\[
48 - t^2 - 16t + 2t^2 = 0
\]
\[
48 + t^2 - 16t = 0
\]
\[
t^2 - 16t + 48 = 0
\]

This is the same equation we obtained in Step 4, so both \( t = 12 \) and \( t = 4 \) are valid solutions.

### Conclusion

The tangent line to the parametric curve passes through the origin at times:

\[
t = 4 \quad \text{and} \quad t = 12
\]

Would you like any further details or clarifications?

---

Here are some related questions to explore:

1. What is the significance of the tangent vector in parametric curves?
2. How do tangent lines help in visualizing curves in space?
3. How does the parameter \( s \) define points on the tangent line?
4. What are other methods to find where a line passes through the origin?
5. How can this approach be applied to more complex parametric curves?

**Tip:** When solving systems of parametric equations, breaking them down one step at a time ensures clarity and prevents mistakes.

Find all times t so that the tangent line to the parametric curve ⟨t − 8, t^2 − 5t − 8, 48 − t^2⟩ at time t passes through the origin

Learn how to find all times t where the tangent line to the parametric curve ⟨t − 8, t^2 − 5t − 8, 48 − t^2⟩ passes through the origin. This solution involves calculating the tangent vector, setting up the tangent line equation, and solving a system of equations. Suitable for students in Grades 10-12.

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