Professor Curious | Talk to an AI Tutor That Explains Until It Clicks

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 $\frac{dx}{dt}$ .

Flip to see [answer/question]

Revise later

SpaceTo flip

If confident

All Flashcards

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.

If the graph of a parametric curve is concave up, what does this indicate about the second derivative?

The second derivative is positive.

How can you visually identify concavity on a graph of a parametric curve?

Concave up: the curve opens upwards. Concave down: the curve opens downwards.

What does a point of inflection on a parametric curve's graph indicate about the second derivative?

The second derivative changes sign at that point.

How does the graph of a cycloid relate to its second derivative?

The graph is always concave down, corresponding to a negative second derivative.

What does a horizontal tangent on a parametric curve indicate about $\frac{dy}{dt}$ ?

$\frac{dy}{dt} = 0$ at that point (assuming $\frac{dx}{dt}$ is not also zero).

What does a vertical tangent on a parametric curve indicate about $\frac{dx}{dt}$ ?

$\frac{dx}{dt} = 0$ at that point (assuming $\frac{dy}{dt}$ is not also zero).

How can you use a graph to estimate the sign of the second derivative at a given point?

Observe the concavity: upward curve suggests positive, downward suggests negative.

If the graph of $\frac{dy}{dx}$ is increasing, what does this imply about $\frac{d^2y}{dx^2}$ ?

$\frac{d^2y}{dx^2}$ is positive.

If the graph of $\frac{dy}{dx}$ is decreasing, what does this imply about $\frac{d^2y}{dx^2}$ ?

$\frac{d^2y}{dx^2}$ is negative.

How does the graph of a parametric equation help visualize the relationship between x, y, and t?

It shows the path traced by the point (x(t), y(t)) as t varies.

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.

All Flashcards

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.