(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the partial derivative with regards to vector r1 for the expression:

theta = acos [tex]\frac{((r1-r2).(r3-r2))}{||r1-r2||*||r3-r2||}[/tex]

where "." is the dot product

r1,r2 and r3 are positions in 3D-space. The expression above comes from the definition of the dot product

2. Relevant equations

r0 = the vector (1,1,1)

r32 = r3-r2

r12 = r1-r2

3. The attempt at a solution

The expression can be rewritten by taking the cosinus of both sides.

Differentiating the left side gives -sin(theta) * [tex]\frac{d theta}{d r1}[/tex]

The right side can be written as f(r) = h(r)/k(r), and which gives the derivative of the form

(h' *k + h *k')/k^2 // Note that I use ' instead of d( )/d r1

h' becomes r0.r32

k' becomes ( || r32|| * (r0.r12) )/||r12||

This gives the right hand side of the equation according to

f ' (r) = [tex]\frac{(r0.r32)*||r12||*||r32||* - \frac{(r12.r32)*||r32||*(r0.r12)}{||r12||}}{(||r12||*||r32||)^2}[/tex]

Thus the original equation becomes

-sin(theta) * d theta/d r1 = f ' (r)

d theta/d r1 = f ' (r)/-sin(theta)

Or alternatively

[tex]\frac{d theta}{d r1}[/tex] = f ' (r)/sqrt(1-f(r)^2)

if we use the derivative of arcos directly. They give the same value.

I've checked my expression for f ' (r) using numerical methods, and it is correct. However, the numerical methods give a different answer than the analytical expression for my final expression of [tex]\frac{d theta}{d r1}[/tex]

Can someone help me figure out what's wrong with my solution?

Thank you for your time!

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# Homework Help: Partial derivative of angle with regards to vector

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