How does $\frac{dx}{dt}$ and $\frac{dy}{dt}$ relate to $\frac{dy}{dx}$ ?

$\frac{dy}{dx}$ is found by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$ .

What does the sign of the second derivative tell you?

Positive: concave up. Negative: concave down. Zero: possible inflection point.

Why is the chain rule important for second derivatives of parametric equations?

It allows us to express the derivative with respect to x in terms of derivatives with respect to t.

How does finding the second derivative of a parametric function relate to finding the first derivative?

Finding the second derivative involves taking the derivative of the first derivative with respect to the parameter t.

Explain the role of the parameter 't' in parametric equations.

The parameter 't' links the x and y coordinates, defining the curve's position at a given 'time'.

What does concavity tell us about the rate of change of the slope?

Concavity describes whether the slope is increasing (concave up) or decreasing (concave down).

What does a negative second derivative mean in the context of a cycloid?

The cycloid is always concave down, meaning it always curves downward.

Explain the steps involved in finding the second derivative of parametric equations.

Find dx/dt and dy/dt, then find dy/dx, then find the derivative of dy/dx with respect to t, and finally divide by dx/dt.

How do you determine the concavity of a parametric curve?

By analyzing the sign of the second derivative, $\frac{d^2y}{dx^2}$ .

Explain the relationship between the first and second derivatives in determining the shape of a parametric curve.

The first derivative gives the slope, while the second derivative gives the concavity, which together define the curve's shape.

Steps to find $\frac{d^2y}{dx^2}$ for $x=t^2$ , $y=t^3$ ?

Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ . 2. Find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ . 3. Find $\frac{d}{dt}(\frac{dy}{dx})$ . 4. Divide by $\frac{dx}{dt}$ .

How to determine where a parametric curve is concave up?

Find $\frac{d^2y}{dx^2}$ . 2. Set $\frac{d^2y}{dx^2} > 0$ . 3. Solve for t.

Find $\frac{d^2y}{dx^2}$ if $x(t) = t^3$ and $y(t) = t^4$ .

$\frac{dx}{dt} = 3t^2$ , $\frac{dy}{dt} = 4t^3$ . 2. $\frac{dy}{dx} = \frac{4}{3}t$ . 3. $\frac{d}{dt}(\frac{dy}{dx}) = \frac{4}{3}$ . 4. $\frac{d^2y}{dx^2} = \frac{4}{9t^2}$ .

How to find the second derivative of $x = t^3 - 3t$ and $y = t^2 + 2t - 5$ ?

Find $\frac{dx}{dt} = 3t^2 - 3$ and $\frac{dy}{dt} = 2t + 2$ . 2. Find $\frac{dy}{dx} = \frac{2}{3(t-1)}$ . 3. Find $\frac{d}{dt}(\frac{dy}{dx}) = -\frac{2}{3(t-1)^2}$ . 4. $\frac{d^2y}{dx^2} = \frac{-2}{9(t-1)(t^2-1)}$ .

Given $x(t) = 2(t - sin(t))$ and $y(t) = 2(1 - cos(t))$ , show the cycloid is concave down.

Find $\frac{dx}{dt} = 2 - 2cos(t)$ and $\frac{dy}{dt} = 2sin(t)$ . 2. Find $\frac{dy}{dx} = \frac{sin(t)}{1 - cos(t)}$ . 3. Find $\frac{d^2y}{dx^2} = \frac{-1}{2(1 - cos(t))^2}$ . 4. Since the second derivative is always negative, the cycloid is always concave down.

How do you simplify $\frac{cos(t)(1-cos(t))-(sin(t))(sin(t))}{(1-cos(t))^2}$ ?

Expand: $\frac{cos(t) - cos^2(t) - sin^2(t)}{(1 - cos(t))^2}$ . 2. Use $sin^2(t) + cos^2(t) = 1$ : $\frac{cos(t) - 1}{(1 - cos(t))^2}$ . 3. Simplify: $\frac{-1}{1 - cos(t)}$ .

What is the first step in finding the second derivative of $x = cos(t)$ , $y = sin(t)$ ?

Find the first derivatives: $\frac{dx}{dt} = -sin(t)$ and $\frac{dy}{dt} = cos(t)$ .

How do you find the slope of the tangent line to a parametric curve at a specific t?

Find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ . 2. Evaluate $\frac{dy}{dx}$ at the given value of t.

How do you find the t-values where a parametric curve has a horizontal tangent?

Find $\frac{dy}{dt}$ . 2. Set $\frac{dy}{dt} = 0$ and solve for t. 3. Ensure $\frac{dx}{dt}$ is not also zero at those t-values.

How do you find the t-values where a parametric curve has a vertical tangent?

Find $\frac{dx}{dt}$ . 2. Set $\frac{dx}{dt} = 0$ and solve for t. 3. Ensure $\frac{dy}{dt}$ is not also zero at those t-values.

Difference between finding $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ for parametric equations?

$\frac{dy}{dx}$ : Find $\frac{dy/dt}{dx/dt}$ . $\frac{d^2y}{dx^2}$ : Find the derivative of $\frac{dy}{dx}$ with respect to t, then divide by $\frac{dx}{dt}$ .

Concave up vs. concave down: second derivative sign?

Concave up: second derivative > 0. Concave down: second derivative < 0.

First derivative vs. second derivative: what do they tell us?

First derivative: slope of the tangent line. Second derivative: concavity.

Parametric equations vs. standard equations: how to find derivatives?

Parametric: use $\frac{dy/dt}{dx/dt}$ . Standard: direct differentiation with respect to x.

Finding horizontal vs. vertical tangents in parametric equations?

Horizontal: set $\frac{dy}{dt} = 0$ . Vertical: set $\frac{dx}{dt} = 0$ .

Quotient rule vs. chain rule: when to use them?

Quotient rule: derivative of a ratio. Chain rule: derivative of a composite function.

$\frac{dy}{dx}$ vs $\frac{dx}{dy}$

$\frac{dy}{dx}$ is the rate of change of y with respect to x, while $\frac{dx}{dy}$ is the rate of change of x with respect to y. They are reciprocals of each other.

First derivative test vs. second derivative test

First derivative test: finds local max/min using sign changes of f'(x). Second derivative test: uses the sign of f''(x) to determine concavity and local extrema.

Explicit vs. implicit differentiation

Explicit: y is directly defined as a function of x (y = f(x)). Implicit: y is defined implicitly in terms of x (e.g., x^2 + y^2 = 1).

Local vs. global extrema

Local: max/min within a specific interval. Global: absolute max/min over the entire domain.

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