To find the slope of the tangent line to the curve defined by parametric equations at a given point $t = 9$, we need to compute $\frac{dy}{dx}$, which is given by:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Step 1: Differentiate both parametric equations with respect to $t$

• $x = 7t^2 + 6t - 2$

Differentiate $x$ with respect to $t$:

$\frac{dx}{dt} = \frac{d}{dt}(7t^2 + 6t - 2) = 14t + 6$

• $y = 2t^3 + 6t^2 - 7$

Differentiate $y$ with respect to $t$:

$\frac{dy}{dt} = \frac{d}{dt}(2t^3 + 6t^2 - 7) = 6t^2 + 12t$

Step 2: Calculate $\frac{dy}{dx}$ at $t = 9$

• First, substitute $t = 9$ into $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

$\frac{dx}{dt} = 14(9) + 6 = 126 + 6 = 132$ $\frac{dy}{dt} = 6(9^2) + 12(9) = 6(81) + 108 = 486 + 108 = 594$

Step 3: Compute the slope

Now, calculate the slope $\frac{dy}{dx}$ by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$:

$\frac{dy}{dx} = \frac{594}{132} \approx 4.500$

Final Answer:

The slope of the tangent line at $t = 9$ is approximately $\boxed{4.500}$.

Would you like further details on this process or have any other questions?

Here are 5 related questions:

1. How do parametric equations differ from Cartesian equations?
2. How would you find the tangent line's equation given a parametric curve?
3. Can parametric equations describe more complex curves than Cartesian forms?
4. What is the significance of second derivatives in parametric curves?
5. How do you convert parametric equations into Cartesian form?

Tip: To find the slope of a parametric curve, always use $\frac{dy/dt}{dx/dt}$, and ensure derivatives are evaluated at the specific parameter value.