Unit 9

Question 1

Arc length

Answer 1

∫ (1 + (dy/dx)^2)^1/2 dx

Question 2

direct distance between two points

Answer 2

ds = ((dx)^2 + (dy)^2)^1/2

ds^2 = dx^2 + dy^2

Question 3

Eliminate the Parameter

Answer 3

1. convert x(t) or y(t) into t = __, then plug into the other
2. use trig. identities to isolate each trig. function

• look up video

Question 4

look up what a vector looks like

Question 5

magnitude/length of a vector

Answer 5

(x^2 + y^2)^1/2

• same as distance formula

Question 6

speed/magnitude of velocity

Answer 6

([x’(t)]^2 + [y’(t)]^2)^1/2

• length but derivatives of x and y

Question 7

the vector from the origin to P is called the`

Answer 7

position vector
- x is the horizontal component
- y is the vertical component

Question 8

the set of position vectors is called

Answer 8

vector function

Question 9

the velocity vector is the

Answer 9

derivative of the vector function

v = <dx/dt, dy/dt>

Question 10

if the vector v is drawn initiating at P, then

Answer 10

it will be tangent to the curve at P and its magnitude with be the speed of the particle at P

Question 11

parametric first derivative

Answer 11

dy/dx = (dy/dt) / (dx/dt)

Question 12

parametric second derivative

Answer 12

d^2y/dx^2 = d/dt (dy/dx) / dx/dt

Question 13

vertical tangent

Answer 13

x’(t) = 0

Question 14

horizontal tangent

Answer 14

y’(t) = 0

Question 15

Make sure you know what the tan, cotx, sec, and csc graphs look like and how to draw them