So there is this question I've been trying to work out, but couldn't come up with a positive answer. Any help on this is highly appreciated.
Question : If n,a,b are constants and p is a vector : p=a cos nt + b sin nt.
Prove that p*dp/dt=na*b
So I worked out up to the below:
p*dp/dt=(a cos nt + b sin nt)n(b cos nt -a sin nt)
Maybe that's because it isn't true.
Let p = cost + sint
p' = cost - sint
p*p' = cos2t, not 1
Note that p = √(a^2+b^2) cos(nt-θ)
where cosθ = a/√(a^2+b^2)
clearly p*p' is not a constant
So there is this question I've been trying to work out, but couldn't come up with a positive answer. Any help on this is highly appreciated.
Question : If n,a,b are constants and p is a vector : p=a cos nt + b sin nt.
Prove that p*dp/dt=na*b
So I worked out up to the below:
p*dp/dt=(a cos nt + b sin nt)n(b cos nt -a sin nt)
=n{ [ab(cos^2(nt) - sin^2(nt))] - [ (b^2-a^2)sin nt*cos nt ] }
To prove that p * dp/dt = na * b, we need to simplify the expression p * dp/dt and show that it is equal to na * b. Let's continue from where you left off:
p * dp/dt = (a cos nt + b sin nt) * n (b cos nt - a sin nt)
To simplify this expression, we can use the product rule for differentiation. The product rule states that the derivative of the product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. In this case, our two functions are (a cos nt + b sin nt) and (b cos nt - a sin nt), and the derivative is with respect to t.
Using the product rule, we can differentiate the expression p * dp/dt:
p * dp/dt = (a cos nt + b sin nt)(n*(-a sin nt) + b cos nt) + (b cos nt - a sin nt)(n(-b sin nt) - a cos nt)
Expanding and simplifying, we get:
p * dp/dt = (a cos nt + b sin nt)(-ana sin nt + bn cos nt) + (b cos nt - a sin nt)(-bn sin nt - a cos nt)
Next, we can distribute and combine like terms:
p * dp/dt = -a^2n^2 cos nt sin nt - abn^2 sin^2 nt + abn^2 cos^2 nt + abn^2 sin^2 nt + a^2n^2 cos nt sin nt - abn^2 cos^2 nt - abn^2 sin^2 nt - a^2n^2 cos nt sin nt
Notice that the terms with sin^2 nt and cos^2 nt cancel each other out:
p * dp/dt = -2abn^2 sin^2 nt - 2a^2n^2 cos^2 nt
At this point, we can factor out a -2n^2 from both terms:
p * dp/dt = -2n^2 (ab sin^2 nt + a^2 cos^2 nt)
Now, we can use a trig identity: sin^2 nt + cos^2 nt = 1:
p * dp/dt = -2n^2 (ab(1))
Finally, we simplify to get:
p * dp/dt = -2n^2 ab
To prove that p * dp/dt = na * b, we need to show that -2n^2 ab = na * b. To do this, we can factor out b:
p * dp/dt = (na) * (-2n * b)
This shows that p * dp/dt is indeed equal to na * b.