Description

Patrick Gardocki
7.1.44
Given: y = −x2 + 3x
Let x = t, y = −t2 + 3t
For point (0,0), t = 0 ⃗r(t) = ti + (−t2 + 3t)j
Tangent vector: drdt = i + (−2t + 3)j
Unit Tangent Vector at point (0,0):

7.2.26
√ √
∥P1,P2∥ = 12 + 22 + 42 = 21
q √ √
∥P1,P3∥ =
q 2 + 02 + (2 2 − 4)2 = p28 − 16 2
∥P2,P3∥ = 2
∥P1,P2∥ = ∥P1,P3∥ ̸= ∥P2,P3∥
∴ isosceles triangle
7.3.45
Given: ∥F∥ = 20, θ = 60◦, ∥d∥ = 100
W = ∥F∥∥d∥cosθ = 20(100)cos(60◦) = 1000 ft-lb
7.3.53
First solve 7.3.52 Prove n = ai + bj is ⊥ to ax + by + c = 0
P1(x1,y1) and P2(x2,y2) are points on the line. P1⃗P2 = (x2−x1,y2−y1) If n·P1⃗P2 = 0 Then perpendicular.
(a,b) · (x2 − x1,y2 − y1) = (ax2 + by2) − (ax1 + by1) = −c − (−c) = 0
∴ n is perpendicular to the line
Prove d c|
d
7.4.52
b × c = 1i 4j k1 = 4 15i − 11 15j + 11 41k = (20 − 1)i − (5 − 1)j + (1 − 4)k = 19i − 4j − 3k

1 1 5 1
v = |a · (b × c)| = (3i + j + k) · 19(i − 4j − 3k) = |57 − 4 − 3| = 50 cu. units
7.5.67
Let z = t 4x − 2y − z = 1; x + y + 2z = 1 → 4x − 2y = 1 + t; x + y = 1 − 2t Solve system of equations. x = − t; y = − t; z = t
7.6.24
p1(x) = x + 1; p2(x) = x − 1
a)
To be linearly independent: c1p1 + c2p2 = 0
c1(x + 1) + c2(x − 1) = (c1 + c2)x + (c1 − c2) = 0 → c1 + c2 = 0; c1 − c2 = 0 ∴ c1 = 0; c2 = 0 b)
p(x) = 5x + 2
c1 + c2 = 5; c1 − c2 = 2 → c
Find Basis, a linearly independent set of vectors that can linearly combine to form vector space:
1 0 0 1 0 0 0
Taking the following set of matrices: ; ; ;
0 0 0 0 1 0 0
These are linearly independent and can represent M22 ∴ a basis.
The dimension is 4 given the 4 matrices used to represent M22
Review Chapter 7.39 0
1
Find line through P = (7,3,−5) that is parallel to = =
d = (4,−2,6) ∴ ∥ line is = =
Review Chapter 7.51
Checking subspace: Pn contains p1 and p2.
d2p1 d2p2 d2p1 d2p2
Vector addition: If = 0, = 0 → = 0
dx2 dx2 dx dx
Scalar Multiplication: Given c as a scalar,= c · 0 = 0
∴ the set of polynomials in Pn satisfying dxd p2 = 0 is a supspace of Pn
Find a basis: Assume a monomial xk, Find dxd22(xk) = 0 d2 (xk) = k(k − 1)xk−2 = 0, When k = 1,x ∴ Basis is {1,x} dx2