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When you're working with parametric equations like \( x = t^3 + 1 \) and \( y = t^2 - t \), finding derivatives involves a few critical steps. In essence, a derivative tells us how a function changes as its input changes, which provides information about the function's rate of change.

For these parametric curves, we first calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). These are derivatives of \( x\) and \( y\) with respect to the parameter \( t\). This is done separately for each function before combining them to find \( \frac{dy}{dx} \).

Here's the breakdown:

• \( \frac{dx}{dt} = 3t^2 \)
• \( \frac{dy}{dt} = 2t - 1 \)

Having these, we get \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t - 1}{3t^2} \), showing how \( y \) changes with \( x \). This formula is central for understanding motion along the curve.

Concavity describes the curvature of a graph. Specifically, it tells us whether a curve bends upwards or downwards. A curve is said to be concave upward if it bends in the shape of a cup (U-shape). In mathematical terms, this happens when the second derivative is positive.

In this problem, concavity is determined by \( \frac{d^2y}{dx^2} \). To glean whether the curve is concave upward, we analyze where \( \frac{d^2y}{dx^2} > 0 \). We found that:

• \( \frac{d^2y}{dx^2} = \frac{2(1-t)}{9t^5} \)

For this expression to be positive, both the top (numerator) and bottom (denominator) must support a positive outcome. Thus, simplifying we get:

• The numerator condition gives \( 1-t > 0 \), or \( t < 1 \).
• The denominator condition demands \( t > 0 \) for positivity.
• Together these conditions give the interval \( 0 < t < 1 \)

This interval specifies where the curve is concave up. It's a critical insight into the shape and nature of this parametric curve.

The second derivative, \( \frac{d^2y}{dx^2} \), tells us about how the rate of change of the slope (first derivative) itself changes. In simpler words, it helps in understanding the *acceleration* of change in a curve. To find it for parametric equations, we need to differentiate \( \frac{dy}{dx} \) with respect to \( t \), then divide by \( \frac{dx}{dt} \):

For \( \frac{dy}{dx} = \frac{2t - 1}{3t^2} \), differentiating with respect to \( t \) yields:

• \( \frac{d}{dt} \left( \frac{2t - 1}{3t^2} \right) = \frac{2(1-t)}{3t^3} \)

After which we divide by \( 3t^2 \) to get:

• \( \frac{d^2y}{dx^2} = \frac{2(1-t)}{9t^5} \).

As seen, the process is more involved than the first derivative, reflecting the added depth of detail it provides about the curve's behavior.

The slope of a tangent at any given point on a curve gives the steepness, or inclination, of the curve at that particular point. When dealing with parametric equations, the slope of the tangent is given by \( \frac{dy}{dx} \).

For the provided parametric equations, the calculated \( \frac{dy}{dx} = \frac{2t - 1}{3t^2} \) represents the instantaneous rate of change of \( y\) with respect to \( x \) for each value of \( t \). This value provides the direction and rate at which the curve is moving.

Understanding the tangent slope is essential for:

• Predicting how the curve behaves at each point.
• Figuring out the points where the curve might have peaks, valleys, or inflections by further analysis with its derivatives.

The slope is a fundamental concept in calculus, illustrating how dynamically curves can evolve along their path.