Calc 3

Question 1

Tangent Line

Answer 1

A line that touches a curve at only 1 point and determines the instantaneous rate of change

Question 2

A line that touches a curve at only 1 point and determines the instantaneous rate of change

Answer 2

Tangent Line

Question 3

r(t) = c(r’(t))

Answer 3

two vectors are parallel

Question 4

two vectors are parallel

Answer 4

r(t) = c(r’(t))

Question 5

Unit Tangent Formula

Answer 5

T(t) = r’(t) / ||r’(t)|| (tangent vector divided by magnitude of tangent vector)

Question 6

T(t) = r’(t) / ||r’(t)|| (tangent vector divided by magnitude of tangent vector)

Answer 6

Unit Tangent Formula

Question 7

Principle Normal Vector

Answer 7

N(t) = T’(t) / ||T’(t)|| ( derivative of unit tangent divided by derivative unit tangent magnitude)

Question 8

N(t) = T’(t) / ||T’(t)|| ( derivative of unit tangent divided by derivative unit tangent magnitude)

Answer 8

Principle Normal Vector

Question 9

Equation for osculating plane

Answer 9

N(t) x T(t) = cross product of Principle Normal Vector and Unit Tangent

Question 10

N(t) x T(t) = cross product of Principle Normal Vector and Unit Tangent

Answer 10

Equation for osculating plane

Question 11

Arc Length Formula

Answer 11

Intergral of : square root ( x’(t) + y’(t) + z’(t) )

Question 12

Intergral of : square root ( x’(t) + y’(t) + z’(t) )

Answer 12

Arc Length Formula

Question 13

||v(t)|| = ||r’(t)||

Answer 13

speed at time t

Question 14

speed at time t

Answer 14

||v(t)|| = ||r’(t)||

Question 15

Curvature of a plane curve

Answer 15

k = |y’’| / ( 1 + (y’)^2 ) ^ 3/2

Question 16

k = |y’’| / ( 1 + (y’)^2 ) ^ 3/2

Answer 16

Curvature of a plane curve

Question 17

r(t) = x(t) + y(t) = given vector function curvature is

Answer 17

k = |x’y’’ - y’x’’| / ( (x’)^2 + (y’)^2 )^ 3/2

Question 18

k = |x’y’’ - y’x’’| / ( (x’)^2 + (y’)^2 )^ 3/2

Answer 18

r(t) = x(t) + y(t) = vector function curvature

Question 19

Curvature of a space curve

Answer 19

k = ||dT/dt|| / (ds/dt)

Question 20

k = ||dT/dt|| / (ds/dt)

Answer 20

Curvature of a space curve

Question 21

(ds/dt)

Answer 21

velocity

Question 22

velocity notation derivative

Answer 22

(ds/dt)

Question 23

aT = Tangential Acceleration

Answer 23

T dot a(t) = depends only on the change of speed

Question 24

T dot a(t) = depends only on the change of speed

Answer 24

aT = Tangential Acceleration

Question 25

aN = Normal component acceleration

Answer 25

|| T x a(t) || = depends on speed and curvature

Question 26

|| T x a(t) || = depends on speed and curvatur

Answer 26

aN = Normal component acceleration

Question 27

curvature of the path of acceleration

Answer 27

k = ||v(t) x a(t)|| / (ds/dt)^3 = cross product magnitude of velocity and acceleration divided by velocity ^ 3

Question 28

k = ||v(t) x a(t)|| / (ds/dt)^3 = cross product magnitude of velocity and acceleration divided by velocity ^ 3

Answer 28

curvature of the path of acceleration

Question 29

Directional Derivative

Answer 29

Gives the rate of change of F in the direction U. | (F) Gradient Vector * (U) unit Vector of Direction

Question 30

Gives the rate of change of F in the direction U. | (F) Gradient Vector * (U) unit Vector of Direction

Answer 30

Directional Derivative

Question 31

Gradient Vector

Answer 31

1st Partial Derivative in respect to (x, y, z)

Question 32

1st Partial Derivative in respect to (x, y, z) forms what vector?

Answer 32

Gradient Vector

Question 33

|| vT(x, y) || = Magnitude of the Directional Derivative implies

Answer 33

Rate of change at (x, y) (rate of increase at (x, y))

Question 34

Rate of change at (x, y) (rate of increase at (x, y))

Answer 34

|| vT(x, y) || = Magnitude of the Directional Derivative

Question 35

- || vT(x, y) || = Magnitude of the Directional Derivative negative implies

Answer 35

Rate of change (x, y) (rate of decrease at (x, y))

Question 36

Rate of change (x, y) (rate of decrease at (x, y))

Answer 36

- || vT(x, y) || = Magnitude of the Directional Derivative negative