Differentiate trigonometric equation

In summary: 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).
  • #1
9988776655
46
0

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

Show your working out.

Homework Equations


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))Answer in book:
(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:
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  • #2
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

Show your working out.

Homework Equations


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))Answer in book:
(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.
 
  • #3
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.
 

Related to Differentiate trigonometric equation

1. What is the basic concept of differentiating a trigonometric equation?

The basic concept of differentiating a trigonometric equation is to find the rate of change or slope of a trigonometric function at any given point. This is done by finding the derivative of the function using the rules of differentiation.

2. What are the common trigonometric functions that can be differentiated?

The common trigonometric functions that can be differentiated include sine, cosine, tangent, cotangent, secant, and cosecant. These functions can be differentiated using the chain rule and other basic rules of differentiation.

3. How do you differentiate a trigonometric equation?

To differentiate a trigonometric equation, you can use the basic rules of differentiation such as the power rule, product rule, quotient rule, and chain rule. These rules are applied to the individual trigonometric functions in the equation to find the derivative.

4. What are the key points to remember when differentiating a trigonometric equation?

When differentiating a trigonometric equation, it is important to remember the basic trigonometric identities and rules, as well as the rules of differentiation. It is also important to carefully apply the chain rule and other rules to each term in the equation.

5. How is differentiation of trigonometric equations used in real life applications?

Differentiation of trigonometric equations is used in various fields such as physics, engineering, and finance. It is used to find the rate of change or slope of a function, which is important in determining the behavior of a system, analyzing data, and making predictions.

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