Question
Use the Chain Rule to find dw/dt.
w = ln (x^2 + y^2 + z^2)^.5
, x = 9 sin t, y = 6 cos t, z = 7 tan t
w = ln (x^2 + y^2 + z^2)^.5
, x = 9 sin t, y = 6 cos t, z = 7 tan t
Answers
Steve
dw/dt = ∂w/∂x dx/dt + ∂w/∂y dy/dt + ∂w/∂z dz/dt
Using ' to mean d/dt,
w = 1/2 ln(x^2+y^2+z^2)
dw/dt = 1/[2(x^2+y^2+z^2)] * (2xx'+2yy'+2zz')
= ((9sin t)(9cos t)+(6cos t)(-6sin t)+(7tan t)(7sec^2 t))/(81sin^2 t + 36cos^2 t + 7tan^2 t)
= (45sint*cost + 49tant*sec^2 t)/(81sin^2 t + 36cos^2 t + 49tan^2 t)
Don't think you can simplify that much.
Using ' to mean d/dt,
w = 1/2 ln(x^2+y^2+z^2)
dw/dt = 1/[2(x^2+y^2+z^2)] * (2xx'+2yy'+2zz')
= ((9sin t)(9cos t)+(6cos t)(-6sin t)+(7tan t)(7sec^2 t))/(81sin^2 t + 36cos^2 t + 7tan^2 t)
= (45sint*cost + 49tant*sec^2 t)/(81sin^2 t + 36cos^2 t + 49tan^2 t)
Don't think you can simplify that much.