1. **Problem:** Given $u = f(2x - 3y, 3y - 4z, 4z - 2x)$, prove that $$\frac{1}{2} \frac{\partial u}{\partial x} + \frac{1}{3} \frac{\partial u}{\partial y} + \frac{1}{4} \frac{\partial u}{\partial z} = 0.$$ **Solution:** Let $p = 2x - 3y$, $q = 3y - 4z$, $r = 4z - 2x$. Then $u = f(p,q,r)$. Use chain rule: $$\frac{\partial u}{\partial x} = f_p \frac{\partial p}{\partial x} + f_q \frac{\partial q}{\partial x} + f_r \frac{\partial r}{\partial x} = f_p \cdot 2 + f_q \cdot 0 + f_r \cdot (-2) = 2f_p - 2f_r,$$ $$\frac{\partial u}{\partial y} = f_p \cdot (-3) + f_q \cdot 3 + f_r \cdot 0 = -3f_p + 3f_q,$$ $$\frac{\partial u}{\partial z} = f_p \cdot 0 + f_q \cdot (-4) + f_r \cdot 4 = -4f_q + 4f_r.$$ Substitute into the identity: $$\frac{1}{2}\frac{\partial u}{\partial x} + \frac{1}{3}\frac{\partial u}{\partial y} + \frac{1}{4}\frac{\partial u}{\partial z} = \frac{1}{2}(2f_p - 2f_r) + \frac{1}{3}(-3f_p + 3f_q) + \frac{1}{4}(-4f_q + 4f_r) = (f_p - f_r) + (-f_p + f_q) + (-f_q + f_r) = 0.$$ Thus proved. 2. **Problem:** State and prove Euler's theorem on homogeneous functions. **Solution:** **Euler's Theorem:** If $f(x,y,\dots)$ is a homogeneous function of degree $n$, then $$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} + \cdots = n f(x,y,\dots).$$ **Proof:** Let $f(tx, ty, \dots) = t^n f(x,y,\dots)$ for all $t$. Differentiate both sides with respect to $t$: $$\frac{d}{dt}f(tx, ty, \dots) = n t^{n-1} f(x,y,\dots).$$ By chain rule: $$\sum \frac{\partial f}{\partial x_i} (x_i) = n t^{n-1} f(x,y,\dots)$$ at $t=1$ this becomes $$\sum x_i \frac{\partial f}{\partial x_i} = n f.$$ This completes the proof. 3a. **Problem:** If $$u = \log(x^2 + y^2) + \tan^{-1} \left( \frac{y}{x} \right),$$ find $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}.$$ **Solution:** Recognize that $u = \log r^2 + \theta$ where $r^2 = x^2 + y^2$ and $\theta = \tan^{-1}(y/x)$. The function $u$ is harmonic, so compute the Laplacian: $$\Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0.$$ This can be verified by direct differentiation or by known results about logarithmic and inverse tangent functions. 3b. **Problem:** Find $$\frac{du}{dt}$$ when $$u = x^2 y + x y^2,$$ with $$x = a t^2$$ and $$y = 2 a t.$$ **Solution:** Use total differentiation: $$\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt}.$$ Calculate partial derivatives: $$\frac{\partial u}{\partial x} = 2 x y + y^2,$$ $$\frac{\partial u}{\partial y} = x^2 + 2 x y.$$ Calculate derivatives: $$\frac{dx}{dt} = 2 a t,$$ $$\frac{dy}{dt} = 2 a.$$ Substitute and simplify: $$\frac{du}{dt} = (2 x y + y^2)(2 a t) + (x^2 + 2 x y)(2 a) = 2 a t (2 x y + y^2) + 2 a (x^2 + 2 x y).$$ Substitute $x = a t^2$, $y = 2 a t$: $$= 2 a t (2 a t^2 \cdot 2 a t + (2 a t)^2) + 2 a (a^2 t^4 + 2 a t^2 \cdot 2 a t) = 2 a t (4 a^2 t^3 + 4 a^2 t^2) + 2 a (a^2 t^4 + 4 a^2 t^3).$$ Simplify powers: $$= 2 a t (4 a^2 t^3 + 4 a^2 t^2) + 2 a (a^2 t^4 + 4 a^2 t^3) = 8 a^3 t^4 + 8 a^3 t^3 + 2 a^3 t^4 + 8 a^3 t^3 = (8 + 2) a^3 t^4 + (8 + 0) a^3 t^3 = 10 a^3 t^4 + 16 a^3 t^3.$$ 4. **Problem:** If $$u = \sin\left(\frac{x}{y}\right), x = e^t, y = t^2,$$ verify $$\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt}.$$ **Solution:** Evaluate LHS: $$\frac{du}{dt} = \cos \left( \frac{x}{y} \right) \cdot \frac{d}{dt} \left( \frac{x}{y} \right).$$ Use quotient rule: $$\frac{d}{dt} \left( \frac{x}{y} \right) = \frac{y \frac{dx}{dt} - x \frac{dy}{dt}}{y^2} = \frac{t^2 e^t - e^t (2 t)}{t^4} = \frac{e^t (t^2 - 2 t)}{t^4}.$$ Evaluate RHS: $$\frac{\partial u}{\partial x} = \cos \left( \frac{x}{y} \right) \frac{1}{y},$$ $$\frac{\partial u}{\partial y} = \cos \left( \frac{x}{y} \right) \left(-\frac{x}{y^2} \right),$$ $$\frac{dx}{dt} = e^t,$$ $$\frac{dy}{dt} = 2 t.$$ So RHS is $$\cos \left( \frac{x}{y} \right) \left( \frac{e^t}{t^2} - \frac{e^t}{t^2} \cdot \frac{2 t}{t^2} \right) = \cos \left( \frac{x}{y} \right) \frac{e^t (t^2 - 2 t)}{t^4}$$ which equals LHS. Hence verified. 5. **Problem:** Given $$r^2 = x^2 + y^2 + z^2$$ and $$V = r^m,$$ prove $$V_{xx} + V_{yy} + V_{zz} = m(m + 1) r^{m-2}.$$ **Solution:** Calculate first derivatives: $$V_x = m r^{m-1} \frac{\partial r}{\partial x} = m r^{m-1} \frac{x}{r} = m x r^{m-2}.$$ Second derivatives: $$V_{xx} = \frac{\partial}{\partial x} (m x r^{m-2}) = m r^{m-2} + m x (m - 2) r^{m-3} \frac{\partial r}{\partial x} = m r^{m-2} + m x (m - 2) r^{m-3} \frac{x}{r} = m r^{m-2} + m (m - 2) x^2 r^{m-4}.$$ Similarly, $$V_{yy} = m r^{m-2} + m (m - 2) y^2 r^{m-4},$$ $$V_{zz} = m r^{m-2} + m (m - 2) z^2 r^{m-4}.$$ Sum: $$V_{xx} + V_{yy} + V_{zz} = 3 m r^{m-2} + m (m - 2) r^{m-4} (x^2 + y^2 + z^2).$$ Note $x^2 + y^2 + z^2 = r^2$, so $$= 3 m r^{m-2} + m (m - 2) r^{m-4} r^2 = 3 m r^{m-2} + m (m - 2) r^{m-2} = m (3 + m - 2) r^{m-2} = m (m + 1) r^{m-2}.$$ Thus proved.