bphct-131-solved-assignment-2024–ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c
BPHCT-131 Solved Assignment 2024 SS
PART A
a) Determine the projection of
A
→
+
2
B
→
A
→
+
2
B
→
vec(A)+2 vec(B) \overrightarrow{\mathbf{A}}+\mathbf{2} \overrightarrow{\mathbf{B}} A → + 2 B →
B
→
B
→
vec(B) \overrightarrow{\mathbf{B}} B →
A
→
=
2
i
^
−
j
^
+
3
k
^
A
→
=
2
i
^
−
j
^
+
3
k
^
vec(A)=2 hat(i)- hat(j)+3 hat(k) \overrightarrow{\mathbf{A}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} A → = 2 i ^ − j ^ + 3 k ^
B
→
=
−
i
^
+
4
j
^
+
k
^
B
→
=
−
i
^
+
4
j
^
+
k
^
vec(B)=- hat(i)+4 hat(j)+ hat(k) \overrightarrow{\mathbf{B}}=-\hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\hat{\mathbf{k}} B → = − i ^ + 4 j ^ + k ^
Answer:
To determine the projection of
A
→
+
2
B
→
A
→
+
2
B
→
vec(A)+2 vec(B) \overrightarrow{\mathbf{A}} + 2\overrightarrow{\mathbf{B}} A → + 2 B → on
B
→
B
→
vec(B) \overrightarrow{\mathbf{B}} B → , we first need to find the vector
A
→
+
2
B
→
A
→
+
2
B
→
vec(A)+2 vec(B) \overrightarrow{\mathbf{A}} + 2\overrightarrow{\mathbf{B}} A → + 2 B → , and then calculate its projection on
B
→
B
→
vec(B) \overrightarrow{\mathbf{B}} B → .
Given vectors:
A
→
=
2
i
^
−
j
^
+
3
k
^
A
→
=
2
i
^
−
j
^
+
3
k
^
vec(A)=2 hat(i)- hat(j)+3 hat(k) \overrightarrow{\mathbf{A}} = 2\hat{\mathbf{i}} – \hat{\mathbf{j}} + 3\hat{\mathbf{k}} A → = 2 i ^ − j ^ + 3 k ^
B
→
=
−
i
^
+
4
j
^
+
k
^
B
→
=
−
i
^
+
4
j
^
+
k
^
vec(B)=- hat(i)+4 hat(j)+ hat(k) \overrightarrow{\mathbf{B}} = -\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + \hat{\mathbf{k}} B → = − i ^ + 4 j ^ + k ^ First, let’s find
A
→
+
2
B
→
A
→
+
2
B
→
vec(A)+2 vec(B) \overrightarrow{\mathbf{A}} + 2\overrightarrow{\mathbf{B}} A → + 2 B → :
A
→
+
2
B
→
=
(
2
i
^
−
j
^
+
3
k
^
)
+
2
(
−
i
^
+
4
j
^
+
k
^
)
A
→
+
2
B
→
=
(
2
i
^
−
j
^
+
3
k
^
)
+
2
(
−
i
^
+
4
j
^
+
k
^
)
vec(A)+2 vec(B)=(2 hat(i)- hat(j)+3 hat(k))+2(- hat(i)+4 hat(j)+ hat(k)) \overrightarrow{\mathbf{A}} + 2\overrightarrow{\mathbf{B}} = (2\hat{\mathbf{i}} – \hat{\mathbf{j}} + 3\hat{\mathbf{k}}) + 2(-\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + \hat{\mathbf{k}}) A → + 2 B → = ( 2 i ^ − j ^ + 3 k ^ ) + 2 ( − i ^ + 4 j ^ + k ^ ) Let’s substitute the values:
=
2
i
^
−
j
^
+
3
k
^
−
2
i
^
+
8
j
^
+
2
k
^
=
2
i
^
−
j
^
+
3
k
^
−
2
i
^
+
8
j
^
+
2
k
^
=2 hat(i)- hat(j)+3 hat(k)-2 hat(i)+8 hat(j)+2 hat(k) = 2\hat{\mathbf{i}} – \hat{\mathbf{j}} + 3\hat{\mathbf{k}} – 2\hat{\mathbf{i}} + 8\hat{\mathbf{j}} + 2\hat{\mathbf{k}} = 2 i ^ − j ^ + 3 k ^ − 2 i ^ + 8 j ^ + 2 k ^ After Calculating we get:
=
(
2
−
2
)
i
^
+
(
−
1
+
8
)
j
^
+
(
3
+
2
)
k
^
=
(
2
−
2
)
i
^
+
(
−
1
+
8
)
j
^
+
(
3
+
2
)
k
^
=(2-2) hat(i)+(-1+8) hat(j)+(3+2) hat(k) = (2 – 2)\hat{\mathbf{i}} + (-1 + 8)\hat{\mathbf{j}} + (3 + 2)\hat{\mathbf{k}} = ( 2 − 2 ) i ^ + ( − 1 + 8 ) j ^ + ( 3 + 2 ) k ^
=
0
i
^
+
7
j
^
+
5
k
^
=
0
i
^
+
7
j
^
+
5
k
^
=0 hat(i)+7 hat(j)+5 hat(k) = 0\hat{\mathbf{i}} + 7\hat{\mathbf{j}} + 5\hat{\mathbf{k}} = 0 i ^ + 7 j ^ + 5 k ^ Now, the projection of a vector
V
→
V
→
vec(V) \overrightarrow{\mathbf{V}} V → on
U
→
U
→
vec(U) \overrightarrow{\mathbf{U}} U → is given by the formula:
proj
U
→
V
→
=
V
→
⋅
U
→
∥
U
→
∥
2
U
→
proj
U
→
V
→
=
V
→
⋅
U
→
∥
U
→
∥
2
U
→
“proj”_( vec(U)) vec(V)=( vec(V)* vec(U))/(|| vec(U)||^(2)) vec(U) \text{proj}_{\overrightarrow{\mathbf{U}}} \overrightarrow{\mathbf{V}} = \frac{\overrightarrow{\mathbf{V}} \cdot \overrightarrow{\mathbf{U}}}{\|\overrightarrow{\mathbf{U}}\|^2} \overrightarrow{\mathbf{U}} proj U → V → = V → ⋅ U → ∥ U → ∥ 2 U → Where
V
→
⋅
U
→
V
→
⋅
U
→
vec(V)* vec(U) \overrightarrow{\mathbf{V}} \cdot \overrightarrow{\mathbf{U}} V → ⋅ U → is the dot product of
V
→
V
→
vec(V) \overrightarrow{\mathbf{V}} V → and
U
→
U
→
vec(U) \overrightarrow{\mathbf{U}} U → , and
∥
U
→
∥
2
∥
U
→
∥
2
|| vec(U)||^(2) \|\overrightarrow{\mathbf{U}}\|^2 ∥ U → ∥ 2 is the magnitude squared of
U
→
U
→
vec(U) \overrightarrow{\mathbf{U}} U → .
Let’s calculate the dot product
V
→
⋅
U
→
V
→
⋅
U
→
vec(V)* vec(U) \overrightarrow{\mathbf{V}} \cdot \overrightarrow{\mathbf{U}} V → ⋅ U → where
V
→
=
0
i
^
+
7
j
^
+
5
k
^
V
→
=
0
i
^
+
7
j
^
+
5
k
^
vec(V)=0 hat(i)+7 hat(j)+5 hat(k) \overrightarrow{\mathbf{V}} = 0\hat{\mathbf{i}} + 7\hat{\mathbf{j}} + 5\hat{\mathbf{k}} V → = 0 i ^ + 7 j ^ + 5 k ^ and
U
→
=
B
→
=
−
i
^
+
4
j
^
+
k
^
U
→
=
B
→
=
−
i
^
+
4
j
^
+
k
^
vec(U)= vec(B)=- hat(i)+4 hat(j)+ hat(k) \overrightarrow{\mathbf{U}} = \overrightarrow{\mathbf{B}} = -\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + \hat{\mathbf{k}} U → = B → = − i ^ + 4 j ^ + k ^ :
V
→
⋅
U
→
=
(
0
)
(
−
1
)
+
(
7
)
(
4
)
+
(
5
)
(
1
)
V
→
⋅
U
→
=
(
0
)
(
−
1
)
+
(
7
)
(
4
)
+
(
5
)
(
1
)
vec(V)* vec(U)=(0)(-1)+(7)(4)+(5)(1) \overrightarrow{\mathbf{V}} \cdot \overrightarrow{\mathbf{U}} = (0)(-1) + (7)(4) + (5)(1) V → ⋅ U → = ( 0 ) ( − 1 ) + ( 7 ) ( 4 ) + ( 5 ) ( 1 ) After Calculating we get:
=
0
+
28
+
5
=
33
=
0
+
28
+
5
=
33
=0+28+5=33 = 0 + 28 + 5 = 33 = 0 + 28 + 5 = 33 Next, calculate
∥
U
→
∥
2
∥
U
→
∥
2
|| vec(U)||^(2) \|\overrightarrow{\mathbf{U}}\|^2 ∥ U → ∥ 2 :
∥
U
→
∥
2
=
(
−
1
)
2
+
(
4
)
2
+
(
1
)
2
∥
U
→
∥
2
=
(
−
1
)
2
+
(
4
)
2
+
(
1
)
2
|| vec(U)||^(2)=(-1)^(2)+(4)^(2)+(1)^(2) \|\overrightarrow{\mathbf{U}}\|^2 = (-1)^2 + (4)^2 + (1)^2 ∥ U → ∥ 2 = ( − 1 ) 2 + ( 4 ) 2 + ( 1 ) 2 After Calculating we get:
=
1
+
16
+
1
=
18
=
1
+
16
+
1
=
18
=1+16+1=18 = 1 + 16 + 1 = 18 = 1 + 16 + 1 = 18 Finally, the projection of
V
→
V
→
vec(V) \overrightarrow{\mathbf{V}} V → on
U
→
U
→
vec(U) \overrightarrow{\mathbf{U}} U → is:
proj
U
→
V
→
=
33
18
U
→
proj
U
→
V
→
=
33
18
U
→
“proj”_( vec(U)) vec(V)=(33)/(18) vec(U) \text{proj}_{\overrightarrow{\mathbf{U}}} \overrightarrow{\mathbf{V}} = \frac{33}{18} \overrightarrow{\mathbf{U}} proj U → V → = 33 18 U → Substituting
U
→
=
−
i
^
+
4
j
^
+
k
^
U
→
=
−
i
^
+
4
j
^
+
k
^
vec(U)=- hat(i)+4 hat(j)+ hat(k) \overrightarrow{\mathbf{U}} = -\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + \hat{\mathbf{k}} U → = − i ^ + 4 j ^ + k ^ :
=
33
18
(
−
i
^
+
4
j
^
+
k
^
)
=
33
18
(
−
i
^
+
4
j
^
+
k
^
)
=(33)/(18)(- hat(i)+4 hat(j)+ hat(k)) = \frac{33}{18}(-\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + \hat{\mathbf{k}}) = 33 18 ( − i ^ + 4 j ^ + k ^ )
=
(
33
18
×
−
1
)
i
^
+
(
33
18
×
4
)
j
^
+
(
33
18
×
1
)
k
^
=
33
18
×
−
1
i
^
+
33
18
×
4
j
^
+
33
18
×
1
k
^
=((33)/(18)xx-1) hat(i)+((33)/(18)xx4) hat(j)+((33)/(18)xx1) hat(k) = \left(\frac{33}{18} \times -1\right)\hat{\mathbf{i}} + \left(\frac{33}{18} \times 4\right)\hat{\mathbf{j}} + \left(\frac{33}{18} \times 1\right)\hat{\mathbf{k}} = ( 33 18 × − 1 ) i ^ + ( 33 18 × 4 ) j ^ + ( 33 18 × 1 ) k ^
=
−
33
18
i
^
+
132
18
j
^
+
33
18
k
^
=
−
33
18
i
^
+
132
18
j
^
+
33
18
k
^
=-(33)/(18) hat(i)+(132)/(18) hat(j)+(33)/(18) hat(k) = -\frac{33}{18}\hat{\mathbf{i}} + \frac{132}{18}\hat{\mathbf{j}} + \frac{33}{18}\hat{\mathbf{k}} = − 33 18 i ^ + 132 18 j ^ + 33 18 k ^ Simplifying:
=
−
11
6
i
^
+
44
6
j
^
+
11
6
k
^
=
−
11
6
i
^
+
44
6
j
^
+
11
6
k
^
=-(11)/(6) hat(i)+(44)/(6) hat(j)+(11)/(6) hat(k) = -\frac{11}{6}\hat{\mathbf{i}} + \frac{44}{6}\hat{\mathbf{j}} + \frac{11}{6}\hat{\mathbf{k}} = − 11 6 i ^ + 44 6 j ^ + 11 6 k ^ Thus, the projection of
A
→
+
2
B
→
A
→
+
2
B
→
vec(A)+2 vec(B) \overrightarrow{\mathbf{A}} + 2\overrightarrow{\mathbf{B}} A → + 2 B → on
B
→
B
→
vec(B) \overrightarrow{\mathbf{B}} B → is
−
11
6
i
^
+
44
6
j
^
+
11
6
k
^
−
11
6
i
^
+
44
6
j
^
+
11
6
k
^
-(11)/(6) hat(i)+(44)/(6) hat(j)+(11)/(6) hat(k) -\frac{11}{6}\hat{\mathbf{i}} + \frac{44}{6}\hat{\mathbf{j}} + \frac{11}{6}\hat{\mathbf{k}} − 11 6 i ^ + 44 6 j ^ + 11 6 k ^ .
b) Obtain the derivative and unit tangent vector at
t
=
1
t
=
1
t=1 t=1 t = 1 for a vector function
a
→
(
t
)
=
t
i
^
+
e
t
2
j
^
+
sin
2
t
k
^
a
→
(
t
)
=
t
i
^
+
e
t
2
j
^
+
sin
2
t
k
^
vec(a)(t)=t hat(i)+e^(t^(2)) hat(j)+sin 2t hat(k) \overrightarrow{\mathbf{a}}(t)=t \hat{\mathbf{i}}+e^{t^2} \hat{\mathbf{j}}+\sin 2 t \hat{\mathbf{k}} a → ( t ) = t i ^ + e t 2 j ^ + sin 2 t k ^ Answer:
To find the derivative of the vector function
a
→
(
t
)
=
t
i
^
+
e
t
2
j
^
+
sin
(
2
t
)
k
^
a
→
(
t
)
=
t
i
^
+
e
t
2
j
^
+
sin
(
2
t
)
k
^
vec(a)(t)=t hat(i)+e^(t^(2)) hat(j)+sin(2t) hat(k) \overrightarrow{\mathbf{a}}(t) = t\hat{\mathbf{i}} + e^{t^2}\hat{\mathbf{j}} + \sin(2t)\hat{\mathbf{k}} a → ( t ) = t i ^ + e t 2 j ^ + sin ( 2 t ) k ^ with respect to
t
t
t t t , we need to differentiate each component of the vector function with respect to
t
t
t t t .
Given:
a
→
(
t
)
=
t
i
^
+
e
t
2
j
^
+
sin
(
2
t
)
k
^
a
→
(
t
)
=
t
i
^
+
e
t
2
j
^
+
sin
(
2
t
)
k
^
vec(a)(t)=t hat(i)+e^(t^(2)) hat(j)+sin(2t) hat(k) \overrightarrow{\mathbf{a}}(t) = t\hat{\mathbf{i}} + e^{t^2}\hat{\mathbf{j}} + \sin(2t)\hat{\mathbf{k}} a → ( t ) = t i ^ + e t 2 j ^ + sin ( 2 t ) k ^ The derivative
d
a
→
d
t
d
a
→
d
t
(d vec(a))/(dt) \frac{d\overrightarrow{\mathbf{a}}}{dt} d a → d t is found by differentiating each component:
For
t
i
^
t
i
^
t hat(i) t\hat{\mathbf{i}} t i ^
d
t
d
t
i
^
=
1
i
^
d
t
d
t
i
^
=
1
i
^
(dt)/(dt) hat(i)=1 hat(i) \frac{dt}{dt}\hat{\mathbf{i}} = 1\hat{\mathbf{i}} d t d t i ^ = 1 i ^
For
e
t
2
j
^
e
t
2
j
^
e^(t^(2)) hat(j) e^{t^2}\hat{\mathbf{j}} e t 2 j ^
d
d
t
(
e
t
2
)
j
^
=
2
t
e
t
2
j
^
d
d
t
(
e
t
2
)
j
^
=
2
t
e
t
2
j
^
(d)/(dt)(e^(t^(2))) hat(j)=2te^(t^(2)) hat(j) \frac{d}{dt}(e^{t^2})\hat{\mathbf{j}} = 2te^{t^2}\hat{\mathbf{j}} d d t ( e t 2 ) j ^ = 2 t e t 2 j ^
For
sin
(
2
t
)
k
^
sin
(
2
t
)
k
^
sin(2t) hat(k) \sin(2t)\hat{\mathbf{k}} sin ( 2 t ) k ^
d
d
t
(
sin
(
2
t
)
)
k
^
=
2
cos
(
2
t
)
k
^
d
d
t
(
sin
(
2
t
)
)
k
^
=
2
cos
(
2
t
)
k
^
(d)/(dt)(sin(2t)) hat(k)=2cos(2t) hat(k) \frac{d}{dt}(\sin(2t))\hat{\mathbf{k}} = 2\cos(2t)\hat{\mathbf{k}} d d t ( sin ( 2 t ) ) k ^ = 2 cos ( 2 t ) k ^
Thus, the derivative of
a
→
(
t
)
a
→
(
t
)
vec(a)(t) \overrightarrow{\mathbf{a}}(t) a → ( t ) is:
d
a
→
d
t
=
1
i
^
+
2
t
e
t
2
j
^
+
2
cos
(
2
t
)
k
^
d
a
→
d
t
=
1
i
^
+
2
t
e
t
2
j
^
+
2
cos
(
2
t
)
k
^
(d vec(a))/(dt)=1 hat(i)+2te^(t^(2)) hat(j)+2cos(2t) hat(k) \frac{d\overrightarrow{\mathbf{a}}}{dt} = 1\hat{\mathbf{i}} + 2te^{t^2}\hat{\mathbf{j}} + 2\cos(2t)\hat{\mathbf{k}} d a → d t = 1 i ^ + 2 t e t 2 j ^ + 2 cos ( 2 t ) k ^ Now, to find the unit tangent vector at
t
=
1
t
=
1
t=1 t=1 t = 1 , we first evaluate the derivative at
t
=
1
t
=
1
t=1 t=1 t = 1 :
d
a
→
d
t
|
t
=
1
=
1
i
^
+
2
(
1
)
e
(
1
)
2
j
^
+
2
cos
(
2
(
1
)
)
k
^
d
a
→
d
t
t
=
1
=
1
i
^
+
2
(
1
)
e
(
1
)
2
j
^
+
2
cos
(
2
(
1
)
)
k
^
(d vec(a))/(dt)|_(t=1)=1 hat(i)+2(1)e^((1)^(2)) hat(j)+2cos(2(1)) hat(k) \left.\frac{d\overrightarrow{\mathbf{a}}}{dt}\right|_{t=1} = 1\hat{\mathbf{i}} + 2(1)e^{(1)^2}\hat{\mathbf{j}} + 2\cos(2(1))\hat{\mathbf{k}} d a → d t | t = 1 = 1 i ^ + 2 ( 1 ) e ( 1 ) 2 j ^ + 2 cos ( 2 ( 1 ) ) k ^
=
1
i
^
+
2
e
j
^
+
2
cos
(
2
)
k
^
=
1
i
^
+
2
e
j
^
+
2
cos
(
2
)
k
^
=1 hat(i)+2e hat(j)+2cos(2) hat(k) = 1\hat{\mathbf{i}} + 2e\hat{\mathbf{j}} + 2\cos(2)\hat{\mathbf{k}} = 1 i ^ + 2 e j ^ + 2 cos ( 2 ) k ^ The unit tangent vector
T
(
t
)
T
(
t
)
T(t) \mathbf{T}(t) T ( t ) is given by:
T
(
t
)
=
d
a
→
d
t
∥
d
a
→
d
t
∥
T
(
t
)
=
d
a
→
d
t
d
a
→
d
t
T(t)=((d vec(a))/(dt))/(||(d vec(a))/(dt)||) \mathbf{T}(t) = \frac{\frac{d\overrightarrow{\mathbf{a}}}{dt}}{\left\|\frac{d\overrightarrow{\mathbf{a}}}{dt}\right\|} T ( t ) = d a → d t ∥ d a → d t ∥ At
t
=
1
t
=
1
t=1 t=1 t = 1 , we need to calculate the magnitude of the derivative vector and then divide the derivative vector by its magnitude to get the unit tangent vector.
The magnitude of the derivative vector at
t
=
1
t
=
1
t=1 t=1 t = 1 is:
∥
d
a
→
d
t
∥
t
=
1
=
(
1
)
2
+
(
2
e
)
2
+
(
2
cos
(
2
)
)
2
d
a
→
d
t
t
=
1
=
(
1
)
2
+
(
2
e
)
2
+
(
2
cos
(
2
)
)
2
||(d vec(a))/(dt)||_(t=1)=sqrt((1)^(2)+(2e)^(2)+(2cos(2))^(2)) \left\|\frac{d\overrightarrow{\mathbf{a}}}{dt}\right\|_{t=1} = \sqrt{(1)^2 + (2e)^2 + (2\cos(2))^2} ∥ d a → d t ∥ t = 1 = ( 1 ) 2 + ( 2 e ) 2 + ( 2 cos ( 2 ) ) 2 Let’s calculate the magnitude and then find the unit tangent vector.
The magnitude of the derivative vector at
t
=
1
t
=
1
t=1 t=1 t = 1 is approximately
5.5901
5.5901
5.5901 5.5901 5.5901 .
Now, to find the unit tangent vector
T
(
t
)
T
(
t
)
T(t) \mathbf{T}(t) T ( t ) at
t
=
1
t
=
1
t=1 t=1 t = 1 , we divide the derivative vector by its magnitude:
T
(
1
)
=
1
i
^
+
2
e
j
^
+
2
cos
(
2
)
k
^
5.5901
T
(
1
)
=
1
i
^
+
2
e
j
^
+
2
cos
(
2
)
k
^
5.5901
T(1)=(1( hat(i))+2e( hat(j))+2cos(2)( hat(k)))/(5.5901) \mathbf{T}(1) = \frac{1\hat{\mathbf{i}} + 2e\hat{\mathbf{j}} + 2\cos(2)\hat{\mathbf{k}}}{5.5901} T ( 1 ) = 1 i ^ + 2 e j ^ + 2 cos ( 2 ) k ^ 5.5901 Let’s calculate each component:
For
i
^
i
^
hat(i) \hat{\mathbf{i}} i ^
1
5.5901
1
5.5901
(1)/(5.5901) \frac{1}{5.5901} 1 5.5901
For
j
^
j
^
hat(j) \hat{\mathbf{j}} j ^
2
e
5.5901
2
e
5.5901
(2e)/(5.5901) \frac{2e}{5.5901} 2 e 5.5901
For
k
^
k
^
hat(k) \hat{\mathbf{k}} k ^
2
cos
(
2
)
5.5901
2
cos
(
2
)
5.5901
(2cos(2))/(5.5901) \frac{2\cos(2)}{5.5901} 2 cos ( 2 ) 5.5901
After Calculating we get:
i
^
i
^
hat(i) \hat{\mathbf{i}} i ^
1
5.5901
≈
0.1789
1
5.5901
≈
0.1789
(1)/(5.5901)~~0.1789 \frac{1}{5.5901} \approx 0.1789 1 5.5901 ≈ 0.1789
j
^
j
^
hat(j) \hat{\mathbf{j}} j ^
2
e
5.5901
≈
0.9802
2
e
5.5901
≈
0.9802
(2e)/(5.5901)~~0.9802 \frac{2e}{5.5901} \approx 0.9802 2 e 5.5901 ≈ 0.9802
k
^
k
^
hat(k) \hat{\mathbf{k}} k ^
2
cos
(
2
)
5.5901
≈
−
0.7554
2
cos
(
2
)
5.5901
≈
−
0.7554
(2cos(2))/(5.5901)~~-0.7554 \frac{2\cos(2)}{5.5901} \approx -0.7554 2 cos ( 2 ) 5.5901 ≈ − 0.7554
Therefore, the unit tangent vector
T
(
1
)
T
(
1
)
T(1) \mathbf{T}(1) T ( 1 ) at
t
=
1
t
=
1
t=1 t=1 t = 1 is approximately:
T
(
1
)
=
0.1789
i
^
+
0.9802
j
^
−
0.7554
k
^
T
(
1
)
=
0.1789
i
^
+
0.9802
j
^
−
0.7554
k
^
T(1)=0.1789 hat(i)+0.9802 hat(j)-0.7554 hat(k) \mathbf{T}(1) = 0.1789\hat{\mathbf{i}} + 0.9802\hat{\mathbf{j}} – 0.7554\hat{\mathbf{k}} T ( 1 ) = 0.1789 i ^ + 0.9802 j ^ − 0.7554 k ^ This vector represents the direction of the curve described by
a
→
(
t
)
a
→
(
t
)
vec(a)(t) \overrightarrow{\mathbf{a}}(t) a → ( t ) at the point where
t
=
1
t
=
1
t=1 t=1 t = 1 , normalized to have a magnitude of 1.