How do we determine the values of t for which the curve defined by x = 2sin(t) and y = 3cos(t) is concave upward?

To analyze the concavity of the given curve, we first need to establish the equations for

dydx and d²ydx². The given parametric equations are:

			 x = 2sin(t) 
			 y = 3cos(t) 

1. **Finding dydt**: We must first find the derivatives of x and y with respect to t:

			 dx/dt = 2cos(t) 
			 dy/dt = -3sin(t) 

2. **Finding dydx**: The derivative
dydx can be found using the chain rule:

			 dy/dx = (dy/dt) / (dx/dt) = (-3sin(t)) / (2cos(t)) = -rac{3}{2}tan(t) 

3. **Finding d²ydx²**: Now, we need to determine d²ydx². We can find it by applying the quotient rule to
dydx:

			 d²y/dx² = d/dt(dy/dx) / (dx/dt) 

Calculating each part:

			 d/dt(dy/dx) = rac{d}{dt}igg(-rac{3}{2}	an(t)igg) = -rac{3}{2}sec²(t) 

So:

			 d²y/dx² = rac{-rac{3}{2}sec²(t)}{2cos(t)} = -rac{3sec²(t)}{4cos(t)} 

4. **Concavity Condition**: The curve is concave upward when d²y/dx² > 0:

			 -rac{3sec²(t)}{4cos(t)} > 0 

Since sec²(t) is always positive, the sign of the expression depends on cos(t). Therefore, we have:

			 cos(t) < 0 

This inequality is satisfied in the intervals where t is in the second quadrant and the third quadrant:

			 t ∈ (π/2, 3π/2) 

Summary: The curve defined by the parametric equations x = 2sin(t) and y = 3cos(t) is concave upward for:

			 t ∈ (π/2, 3π/2) 

That is, for the range of values of t between π/2 and 3π/2, the curve exhibits concave upward behavior.

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