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.