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To find the equation of the tangent to the curve at the given point, let's go through the steps. We have the parametric equations:

\[
x = t - t^{-1} \quad \text{and} \quad y = 7 + t^2
\]
at \( t = 1 \).

### Step 1: Find the point on the curve at \( t = 1 \)
Substitute \( t = 1 \) into the equations for \( x \) and \( y \):

\[
x = 1 - 1^{-1} = 1 - 1 = 0
\]
\[
y = 7 + 1^2 = 7 + 1 = 8
\]

So, the point on the curve when \( t = 1 \) is \( (0, 8) \).

### Step 2: Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)
Calculate the derivatives of \( x \) and \( y \) with respect to \( t \):

\[
\frac{dx}{dt} = 1 + t^{-2} = 1 + \frac{1}{t^2}
\]
\[
\frac{dy}{dt} = 2t
\]

### Step 3: Evaluate the derivatives at \( t = 1 \)
Now substitute \( t = 1 \) into \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):

\[
\frac{dx}{dt}\Big|_{t=1} = 1 + \frac{1}{1^2} = 1 + 1 = 2
\]
\[
\frac{dy}{dt}\Big|_{t=1} = 2 \times 1 = 2
\]

### Step 4: Find the slope of the tangent line \(\frac{dy}{dx}\)
The slope of the tangent line is given by:

\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{2} = 1
\]

### Step 5: Write the equation of the tangent line
Using the point-slope form of a line, \( y - y_1 = m(x - x_1) \), with \( m = 1 \) and the point \( (0, 8) \):

\[
y - 8 = 1(x - 0)
\]
\[
y = x + 8
\]

### Final Answer:
The equation of the tangent line to the curve at \( t = 1 \) is:

\[
y = x + 8
\]

Would you like further details or have any questions?

Here are some related questions to consider:

1. How would the tangent line change if the parameter \( t \) was different?
2. What are the geometric interpretations of the parametric derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)?
3. How does the tangent line relate to the rate of change of the curve at a point?
4. Could we use implicit differentiation if the curve wasn't in parametric form?
5. How does changing the parametric equations affect the curve's shape and tangent properties?

**Tip:** Always double-check your derivative calculations, especially with parametric equations, as errors there can lead to incorrect slopes.

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter: x = t - t^(-1), y = 7 + t^2, t = 1.

Learn how to find the equation of a tangent to a curve defined by parametric equations x = t - t^(-1) and y = 7 + t^2 at the point where t = 1. This solution involves derivatives of parametric equations and the point-slope form of a line. Ideal for students in Grades 11-12.

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