To find $\frac{dy}{dx}$ for the parametric curve defined by: $x = 3 \sin t \quad \text{and} \quad y = 2 \cos t$

we need to calculate the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$ with respect to $t$ and then use the formula: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.$

Step 1: Differentiate $x$ and $y$ with respect to $t$

1. Differentiate $x = 3 \sin t$: $\frac{dx}{dt} = 3 \cos t.$

2. Differentiate $y = 2 \cos t$: $\frac{dy}{dt} = -2 \sin t.$

Step 2: Substitute into $\frac{dy}{dx}$

Now, we substitute these derivatives into the formula: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2 \sin t}{3 \cos t} = -\frac{2}{3} \tan t.$

Slope of the Curve at the Indicated Point

To find the slope at a specific point, substitute the value of $t$ that corresponds to that point into $-\frac{2}{3} \tan t$. If you provide the specific point, I can calculate the exact slope for you.

Would you like details on the specific point to continue?

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Here are five follow-up questions to extend this information:

1. How do we interpret the result of $\frac{dy}{dx}$ geometrically on a parametric curve?
2. How would we proceed if the parametric equations were given in terms of a different parameter?
3. What if we wanted to find $\frac{d^2y}{dx^2}$ for this parametric curve?
4. How does the approach differ if the curve was defined implicitly instead of parametrically?
5. What is the effect on $\frac{dy}{dx}$ if the coefficients in $x = a \sin t$ and $y = b \cos t$ were different?

Tip: Always check for periodic points on parametric curves like sine and cosine to simplify calculations for specific values.