# Derivatives of Vectors Lesson 10.4 ```Derivatives of Vectors
Lesson 10.4
Component Vectors
• Unit vectors often used to express vectors

P = Pxi + Py j
 i and j are vectors with length 1, parallel to x
and y axes, respectively
P = Pxi + Py j
j
i
2
Vector Functions and
Parametric Equations
• Consider a curve described by parametric
equations

x = f(t)
y = g(t)
• The curve can be expressed as the
vector-valued function, P(t)

P(t) = f(t)i + g(t)j
t =t 1= 2
t=3
t = 4t = 5
3
Example
• Consider the curve represented by
parametric equations
x  ln t 2
y  cos 1  t 2
• Then the vector-valued function is …
P(t )  ln t i  cos 1  t j
2
2
4
Derivatives of Vector-Valued
Functions
• Given the vector valued function
p(t) = f(t)i + g(t)j

Given also that f(t) and g(t) are differentiable
• Then the derivative of p is
p'(t) = f '(t)i + g'(t)j
• Recall that if p is a position function

p'(t) is the velocity function
 p''(t) is the acceleration function
5
Example
• Given parametric equations which
describe a vector-valued position function
x = t3 – t
 y = 4t – 3t2

• What is the velocity vector?
v(t )   3t 2  1 i   4  6t  j
• What is the acceleration vector?
a(t )   6t  i   6 j
6
Example
• For the same vector-valued function

x = t3 – t
and
y = 4t – 3t2
• What is the magnitude of v(t) when t = 1?
v (t )   3t 2  1 i   4  6t  j
v (1)  2i  2 j
v  22  22  8  2 2
• The direction?

 2  7
tan   

4
 2  4
1
7
Application
• The Easter Bunny is traveling by balloon

Position given by height y = 360t – 9t2 and
x = 0.8t2 + 0.9 sin2t (positive direction west)
• Determine the velocity of the balloon at
any time t
• For time t = 2.5, determine

Position
 Speed
 Direction
8
Assignment
• Lesson 10.4
• Page 426
• Exercises 1 – 13 odd
9
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