find dy/dx and the slope of the curve at the indicated point. Which of the following gives dy/dx for the
parametric curve x ! 3 sin t, y ! 2 cos t?

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.

find dy/dx and the slope of the curve at the indicated point. Which of the following gives dy/dx for the parametric curve x = 3 sin t, y = 2 cos t?

Discover how to find dy/dx and the slope of the parametric curve x = 3 sin t, y = 2 cos t. This detailed solution explains parametric differentiation, using calculus concepts suitable for students in Grades 11-12.

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