Questions: μ = x^4 y + y^2 z^3 x = r s e^t y = r s^2 e^-t z = r^2 s sin t find ∂u/∂s when r = α, Δ = 1 and t = 0 (sin x)' = cos x (cos x)' = -sin x sin 0 = 0 cos 0 = 1

Transcript text: \[ \mu=x^{4} y+y^{2} z^{3} \quad \begin{array}{l} x=r s e^{t} \\ y=r s^{2} e^{-t} \\ z=r^{2} s \sin t \end{array} \] find $\frac{\partial u}{\partial s}$ when $r=\alpha, \Delta=1$ and $t=0$ \[ \begin{array}{l} (\sin x)^{\prime}=\cos x \\ (\cos x)^{\prime}=-\sin x \\ \sin 0=0 \\ \cos 0=1 \end{array} \]

Solution

Solution Steps

To find $\frac{\partial \mu}{\partial s}$, we need to use the chain rule and substitute the given expressions for $x$, $y$, and $z$. Then, we evaluate the partial derivative at the specified values of $r$, $\Delta$, and $t$.

Substitute $x$, $y$, and $z$ into $\mu$.
Compute the partial derivative $\frac{\partial \mu}{\partial s}$.
Evaluate the derivative at $r = \alpha$, $\Delta = 1$, and $t = 0$.

Step 1: Substitute $x$, $y$, and $z$ into $\mu$

Given: \[ \mu = x^4 y + y^2 z^3 \] Substitute: \[ x = r s e^t, \quad y = r s^2 e^{-t}, \quad z = r^2 s \sin t \] into $\mu$: \[ \mu = (r s e^t)^4 (r s^2 e^{-t}) + (r s^2 e^{-t})^2 (r^2 s \sin t)^3 \] Simplify: \[ \mu = r^5 s^6 e^{3t} + r^8 s^7 e^{-2t} \sin^3 t \]

Step 2: Compute the partial derivative $\frac{\partial \mu}{\partial s}$

Differentiate $\mu$ with respect to $s$: \[ \frac{\partial \mu}{\partial s} = \frac{\partial}{\partial s} \left( r^5 s^6 e^{3t} + r^8 s^7 e^{-2t} \sin^3 t \right) \] \[ \frac{\partial \mu}{\partial s} = 6 r^5 s^5 e^{3t} + 7 r^8 s^6 e^{-2t} \sin^3 t \]

Step 3: Evaluate the derivative at $r = \alpha$, $t = 0$

Substitute $r = \alpha$ and $t = 0$ into $\frac{\partial \mu}{\partial s}$: \[ \frac{\partial \mu}{\partial s} \bigg|_{r = \alpha, t = 0} = 6 \alpha^5 s^5 e^{3 \cdot 0} + 7 \alpha^8 s^6 e^{-2 \cdot 0} \sin^3 0 \] Since $\sin 0 = 0$ and $e^0 = 1$: \[ \frac{\partial \mu}{\partial s} \bigg|_{r = \alpha, t = 0} = 6 \alpha^5 s^5 \]