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  1. AP Calculus
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Define parametric equations.

Functions where independent functions x and y are connected via a parameter t.

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Define parametric equations.

Functions where independent functions x and y are connected via a parameter t.

What is a second derivative?

The derivative of the first derivative of a function.

What is d2ydx2\frac{d^2y}{dx^2}dx2d2y​?

Notation for the second derivative of y with respect to x.

Define concavity.

The direction of the curve of a function (upward or downward).

What is a cycloid?

A curve traced by a point on a circle as it rolls along a straight line.

Define dydx\frac{dy}{dx}dxdy​ for parametric equations.

The first derivative of y with respect to x, found by dy/dtdx/dt\frac{dy/dt}{dx/dt}dx/dtdy/dt​.

What does the second derivative indicate?

The rate of change of the slope of a function; indicates concavity.

What is the quotient rule?

A rule for finding the derivative of a function that is the ratio of two other functions.

What is a parameter?

An independent variable (often denoted by 't') that relates two dependent variables.

What does it mean for a curve to be concave down?

The curve bends downwards, and its second derivative is negative.

Steps to find d2ydx2\frac{d^2y}{dx^2}dx2d2y​ for x=t2x=t^2x=t2, y=t3y=t^3y=t3?

  1. Find dxdt\frac{dx}{dt}dtdx​ and dydt\frac{dy}{dt}dtdy​. 2. Find dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}dxdy​=dx/dtdy/dt​. 3. Find ddt(dydx)\frac{d}{dt}(\frac{dy}{dx})dtd​(dxdy​). 4. Divide by dxdt\frac{dx}{dt}dtdx​.

How to determine where a parametric curve is concave up?

  1. Find d2ydx2\frac{d^2y}{dx^2}dx2d2y​. 2. Set d2ydx2>0\frac{d^2y}{dx^2} > 0dx2d2y​>0. 3. Solve for t.

Find d2ydx2\frac{d^2y}{dx^2}dx2d2y​ if x(t)=t3x(t) = t^3x(t)=t3 and y(t)=t4y(t) = t^4y(t)=t4.

  1. dxdt=3t2\frac{dx}{dt} = 3t^2dtdx​=3t2, dydt=4t3\frac{dy}{dt} = 4t^3dtdy​=4t3. 2. dydx=43t\frac{dy}{dx} = \frac{4}{3}tdxdy​=34​t. 3. ddt(dydx)=43\frac{d}{dt}(\frac{dy}{dx}) = \frac{4}{3}dtd​(dxdy​)=34​. 4. d2ydx2=49t2\frac{d^2y}{dx^2} = \frac{4}{9t^2}dx2d2y​=9t24​.

How to find the second derivative of x=t3−3tx = t^3 - 3tx=t3−3t and y=t2+2t−5y = t^2 + 2t - 5y=t2+2t−5?

  1. Find dxdt=3t2−3\frac{dx}{dt} = 3t^2 - 3dtdx​=3t2−3 and dydt=2t+2\frac{dy}{dt} = 2t + 2dtdy​=2t+2. 2. Find dydx=23(t−1)\frac{dy}{dx} = \frac{2}{3(t-1)}dxdy​=3(t−1)2​. 3. Find ddt(dydx)=−23(t−1)2\frac{d}{dt}(\frac{dy}{dx}) = -\frac{2}{3(t-1)^2}dtd​(dxdy​)=−3(t−1)22​. 4. d2ydx2=−29(t−1)(t2−1)\frac{d^2y}{dx^2} = \frac{-2}{9(t-1)(t^2-1)}dx2d2y​=9(t−1)(t2−1)−2​.

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

  1. Find dxdt=2−2cos(t)\frac{dx}{dt} = 2 - 2cos(t)dtdx​=2−2cos(t) and dydt=2sin(t)\frac{dy}{dt} = 2sin(t)dtdy​=2sin(t). 2. Find dydx=sin(t)1−cos(t)\frac{dy}{dx} = \frac{sin(t)}{1 - cos(t)}dxdy​=1−cos(t)sin(t)​. 3. Find d2ydx2=−12(1−cos(t))2\frac{d^2y}{dx^2} = \frac{-1}{2(1 - cos(t))^2}dx2d2y​=2(1−cos(t))2−1​. 4. Since the second derivative is always negative, the cycloid is always concave down.

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

  1. Expand: cos(t)−cos2(t)−sin2(t)(1−cos(t))2\frac{cos(t) - cos^2(t) - sin^2(t)}{(1 - cos(t))^2}(1−cos(t))2cos(t)−cos2(t)−sin2(t)​. 2. Use sin2(t)+cos2(t)=1sin^2(t) + cos^2(t) = 1sin2(t)+cos2(t)=1: cos(t)−1(1−cos(t))2\frac{cos(t) - 1}{(1 - cos(t))^2}(1−cos(t))2cos(t)−1​. 3. Simplify: −11−cos(t)\frac{-1}{1 - cos(t)}1−cos(t)−1​.

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

Find the first derivatives: dxdt=−sin(t)\frac{dx}{dt} = -sin(t)dtdx​=−sin(t) and dydt=cos(t)\frac{dy}{dt} = cos(t)dtdy​=cos(t).

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

  1. Find dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}dxdy​=dx/dtdy/dt​. 2. Evaluate dydx\frac{dy}{dx}dxdy​ at the given value of t.

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

  1. Find dydt\frac{dy}{dt}dtdy​. 2. Set dydt=0\frac{dy}{dt} = 0dtdy​=0 and solve for t. 3. Ensure dxdt\frac{dx}{dt}dtdx​ is not also zero at those t-values.

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

  1. Find dxdt\frac{dx}{dt}dtdx​. 2. Set dxdt=0\frac{dx}{dt} = 0dtdx​=0 and solve for t. 3. Ensure dydt\frac{dy}{dt}dtdy​ is not also zero at those t-values.

Difference between finding dydx\frac{dy}{dx}dxdy​ and d2ydx2\frac{d^2y}{dx^2}dx2d2y​ for parametric equations?

dydx\frac{dy}{dx}dxdy​: Find dy/dtdx/dt\frac{dy/dt}{dx/dt}dx/dtdy/dt​. d2ydx2\frac{d^2y}{dx^2}dx2d2y​: Find the derivative of dydx\frac{dy}{dx}dxdy​ with respect to t, then divide by dxdt\frac{dx}{dt}dtdx​.

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 dy/dtdx/dt\frac{dy/dt}{dx/dt}dx/dtdy/dt​. Standard: direct differentiation with respect to x.

Finding horizontal vs. vertical tangents in parametric equations?

Horizontal: set dydt=0\frac{dy}{dt} = 0dtdy​=0. Vertical: set dxdt=0\frac{dx}{dt} = 0dtdx​=0.

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

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

dydx\frac{dy}{dx}dxdy​ vs dxdy\frac{dx}{dy}dydx​

dydx\frac{dy}{dx}dxdy​ is the rate of change of y with respect to x, while dxdy\frac{dx}{dy}dydx​ 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.