MATH2310 – MVC Lecture 1
Chain rule:
Differentiate the outside, multiply by the derivative of the inside.
If z=f ( x , y ), and x=g(t), y=h(t). Then:
dz ∂ z dx ∂ z dy
= +
dt ∂ x dt ∂ y dt
If z=f ( x , y ), and x=g(s , t), y=h(s , t). Then:
∂z ∂z ∂x ∂ z ∂ y ∂z ∂z ∂x ∂ z ∂ y
= + , and = +
∂s ∂ x ∂ s ∂ y ∂s ∂t ∂ x ∂t ∂ y ∂t
Function gradient:
The gradient of a bivariate function is given by the vector:
∇f ≔ ⟨ ∂f ∂f
,
∂x ∂y ⟩
Polar coordinates:
A polar coordinate is given by a length and an angle, (r , θ).
A cartesian coordinate can be converted to polar using basic trig; the conversion
looks like:
(√ x 2+ y 2 , tan −1 ( xy ))
Cylindrical coordinates:
Cylindrical coordinates are like polar coordinates with height, (r , θ , z).
To convert to and from cartesian, use the following conversions:
(r , θ , z)←
(√ x 2 + y 2 , tan−1 ( yx ) , z)
(x , y , z)←(r cos θ , sin θ , z )
Chain rule:
Differentiate the outside, multiply by the derivative of the inside.
If z=f ( x , y ), and x=g(t), y=h(t). Then:
dz ∂ z dx ∂ z dy
= +
dt ∂ x dt ∂ y dt
If z=f ( x , y ), and x=g(s , t), y=h(s , t). Then:
∂z ∂z ∂x ∂ z ∂ y ∂z ∂z ∂x ∂ z ∂ y
= + , and = +
∂s ∂ x ∂ s ∂ y ∂s ∂t ∂ x ∂t ∂ y ∂t
Function gradient:
The gradient of a bivariate function is given by the vector:
∇f ≔ ⟨ ∂f ∂f
,
∂x ∂y ⟩
Polar coordinates:
A polar coordinate is given by a length and an angle, (r , θ).
A cartesian coordinate can be converted to polar using basic trig; the conversion
looks like:
(√ x 2+ y 2 , tan −1 ( xy ))
Cylindrical coordinates:
Cylindrical coordinates are like polar coordinates with height, (r , θ , z).
To convert to and from cartesian, use the following conversions:
(r , θ , z)←
(√ x 2 + y 2 , tan−1 ( yx ) , z)
(x , y , z)←(r cos θ , sin θ , z )
