# Differentiate trigonometric equation

• 9988776655
But I am not sure what to do with (C*r). I believe that I need to use the product rule on (C*r) but I am not sure how to do that with a matrix.In summary, the problem involves finding the derivatives of the function y with respect to θ, Φ, and ψ. The function y is equal to (Ua - Ub)' * C * r, where C is a 3x3 rotation matrix and r, Ua, and Ub are 3x1 column vectors. To find the derivatives, the product rule can be used on (C*r).

## Homework Statement

a) Differentiate the following equation with respect to:
1) θ
2) Φ
3) ψ

(Ua - Ub)' * C * r
where:

C is a 3 x 3 rotation matrix:
[ cos θ cos ψ, -cos Φ sin ψ + sin Φ sin θ cos ψ, sin Φ sin ψ + cos Φ sin θ cos ψ]
[ cos θ sin ψ, cos Φ cos ψ + sin Φ sin θ sin ψ, -sin Φ cos ψ + cos Φ sin θ sin ψ]
[ -sin θ, sin Φ cos θ, cos Φ cos θ ]

Ua is a 3x1 column vector:
[Ua_x]
[Ua_y]
[Ua_z]
Ub is a 3x1 column vector:
[Ub_x]
[Ub_y]
[Ub_z]
r is a 3 x 1 column vector:
[r_x]
[r_y]
[r_z]
' means transpose

derivative of:
sin x is cos x
cos x is -sin x

## The Attempt at a Solution

Let:
θ = theta
Φ = phi
ψ = psi

Expanding:
r_z*((Ua_x - Ub_x)*(sin(phi)*sin(psi) + cos(phi)*cos(psi)*sin(theta)) - (Ua_y - Ub_y)*(cos(psi)*sin(phi) - cos(phi)*sin(psi)*sin(theta)) + cos(phi)*cos(theta)*(Ua_z - Ub_z)) + r_y*((Ua_y - Ub_y)*(cos(phi)*cos(psi) + sin(phi)*sin(psi)*sin(theta)) - (Ua_x - Ub_x)*(cos(phi)*sin(psi) - cos(psi)*sin(phi)*sin(theta)) + cos(theta)*sin(phi)*(Ua_z - Ub_z)) + r_x*(cos(psi)*cos(theta)*(Ua_x - Ub_x) - sin(theta)*(Ua_z - Ub_z) + cos(theta)*sin(psi)*(Ua_y - Ub_y))

(Ua - Ub)' * (skew(C * r))
Where skew is the skew symmetric matrix
ie skew (x y z) =
[0 -z y]
[z 0 -x]
[-y x 0]

Last edited:
9988776655 said:

## Homework Statement

a) Differentiate the following equation with respect to:
1) θ
2) Φ
3) ψ

(Ua - Ub)' * C * r
The above is not an equation -- no = in it.
9988776655 said:
where:

C is a 3 x 3 rotation matrix:
[ cos θ cos ψ, -cos Φ sin ψ + sin Φ sin θ cos ψ, sin Φ sin ψ + cos Φ sin θ cos ψ]
[ cos θ sin ψ, cos Φ cos ψ + sin Φ sin θ sin ψ, -sin Φ cos ψ + cos Φ sin θ sin ψ]
[ -sin θ, sin Φ cos θ, cos Φ cos θ ]

Ua is a 3x1 column vector:
[Ua_x]
[Ua_y]
[Ua_z]
Ub is a 3x1 column vector:
[Ub_x]
[Ub_y]
[Ub_z]
r is a 3 x 1 column vector:
[r_x]
[r_y]
[r_z]
' means transpose

derivative of:
sin x is cos x
cos x is -sin x

## The Attempt at a Solution

Let:
θ = theta
Φ = phi
ψ = psi

Expanding:
r_z*((Ua_x - Ub_x)*(sin(phi)*sin(psi) + cos(phi)*cos(psi)*sin(theta)) - (Ua_y - Ub_y)*(cos(psi)*sin(phi) - cos(phi)*sin(psi)*sin(theta)) + cos(phi)*cos(theta)*(Ua_z - Ub_z)) + r_y*((Ua_y - Ub_y)*(cos(phi)*cos(psi) + sin(phi)*sin(psi)*sin(theta)) - (Ua_x - Ub_x)*(cos(phi)*sin(psi) - cos(psi)*sin(phi)*sin(theta)) + cos(theta)*sin(phi)*(Ua_z - Ub_z)) + r_x*(cos(psi)*cos(theta)*(Ua_x - Ub_x) - sin(theta)*(Ua_z - Ub_z) + cos(theta)*sin(psi)*(Ua_y - Ub_y))

(Ua - Ub)' * (skew(C * r))
Where skew is the skew symmetric matrix
ie skew (x y z) =
[0 -z y]
[z 0 -x]
[-y x 0]
The problem isn't clear to me. From your problem statement, are you supposed to find three separate derivatives or are you supposed to find the derivative with respect to θ, and then differentiate that function with respect to Φ, and then, finally, differentiate that function with respect to ψ?

As I understand this problem, expanding the original expression as you did makes things worse. ##U_a## and ##U_b## don't involve θ, Φ, or ψ, so ##(U_a - U_b)^T## can be treated as a constant, using the differentiation rule ##\frac d {dx}(k f(x)) = k \frac d {dx} f(x)##. Also, r doesn't appear to involve θ, Φ, or ψ, so differentiating Cr would be a lot simpler.

Hi,

Sorry for not being specific enough. You can write
y = (Ua - Ub)' * C * r;
then find
1) ∂y/∂θ
2) ∂y/∂Φ
3) ∂y/∂ψ
so find three separate derivatives.
Yes I see that (Ua - Ub)' is constant.