Vector Functions

Question

r(t)=< e^5t cos(12t), e^5tsin(12t), e^5t >

 

1. Show that (5cos12t - 12sin12t)² + (5sin12t + 12cos12t)² = 13²

 

2. Find the equation of the surface the curve is on and name the surface?

 

3. Find the velocity and speed

 

4. Find the unit tangent vector T(t)

 

5. Find the unit normal vector N(t)

 

 

 

Answer 1

1) Show the identity

r(t) = < e^(5t)cos(12t), e^(5t)sin(12t), e^(5t) >

(5cos(12t) − 12sin(12t))^2 + (5sin(12t) + 12cos(12t))^2

= 25cos^2(12t) − 120sin(12t)cos(12t) + 144sin^2(12t)

+ 25sin^2(12t) + 120sin(12t)cos(12t) + 144cos^2(12t)

Cross terms cancel

= (25 + 144)(cos^2(12t) + sin^2(12t))

= 169(1)= 169

2) Equation and name of the surface

x = e^(5t)cos(12t)

y = e^(5t)sin(12t)

z = e^(5t)

x^2 + y^2 = e^(10t)(cos^2(12t) + sin^2(12t))

x^2 + y^2 = e^(10t)

Since z = e^(5t), then z^2 = e^(10t)

x^2 + y^2 = z^2

The curve lies on a right circular cone

3) Velocity and speed

v(t) = r'(t)

v(t) = < e^(5t)(5cos(12t) − 12sin(12t)),

e^(5t)(5sin(12t) + 12cos(12t)),

5e^(5t) >

|v(t)| = sqrt[(e^(5t))^2((5cos(12t) − 12sin(12t))^2

+ (5sin(12t) + 12cos(12t))^2 + 25)]

|v(t)| = e^(5t)sqrt(169 + 25)

|v(t)| = e^(5t)sqrt(194)

4) Unit tangent vector T(t)

T(t) = v(t) / |v(t)|

T(t) = (1 / sqrt(194)) <

5cos(12t) − 12sin(12t),

5sin(12t) + 12cos(12t),

5 >

5) Unit normal vector N(t)

T'(t) = (12 / sqrt(194)) <

−5sin(12t) − 12cos(12t),

5cos(12t) − 12sin(12t),

0 >

|T'(t)| = 156 / sqrt(194)

N(t) = T'(t) / |T'(t)|

N(t) = (1 / 13) <

−5sin(12t) − 12cos(12t),

5cos(12t) − 12sin(12t),

0 >