Question

If w(x, y, z) = x² + y² + z², x = e^t, y = e^t sin t and z = e^t cos t, find ${\displaystyle \frac{\text{d}w}{\text{dt}}}$

Answer 1

w(x, y, z) = x² + y² + z², x = e^t, y = e^t sin t and z = e^t cos t

${\displaystyle \frac{\text{d}w}{\text{dt}}=\frac{\text{d}w}{\text{d}x}\frac{\text{d}x}{\text{dt}}+\frac{\text{d}w}{\text{d}y}\frac{\text{d}y}{\text{dt}}+\frac{\text{d}w}{\text{dt}}+\frac{\text{d}w}{\text{d}z}\frac{\text{d}z}{\text{dt}}}$

${\displaystyle \frac{\text{d}w}{\text{d}x}=2x,\frac{\text{d}x}{\text{dt}}={\text{e}}^{\text{t}}}$

${\displaystyle \frac{\text{d}w}{\text{d}y}=2y,\frac{\text{d}y}{\text{dt}}={\text{e}}^{\text{t}}\sin \text{t}+{\text{e}}^{\text{t}}\cos \text{t}}$

${\displaystyle \frac{\text{d}w}{\text{d}z}=2z,\frac{\text{d}z}{\text{dt}}={\text{e}}^{\text{t}}\cos \text{t}-{\text{e}}^{\text{t}}\sin \text{t}}$

${\displaystyle \frac{\text{d}w}{\text{dt}}}$ = 2x e^t + 2y (e^t sin t + e^t cos t) + 2z (e^t cos t – e^t sin t)

= 2 e^2t + 2 (e^t sin t) (e^t sin t + e^t cos t) + 2 (e^t cos t) (e^t cos t – e^t sin t)

= 2 e^2t [1 + sin²t + sin t cos t + cos²t – sin t cos t]

= 2 e^2t (1 + sin²t + cos²t)  ......[∵ sin²t + cos²t = 1]

= 2 e²t (1 + 1)

= 4 e^2t