1. **Problem Statement:** Given the vector function $$\mathbf{w} = \begin{bmatrix} 4t \\ r \\ s \end{bmatrix}$$ where $$t = \tan^{-1}\left(\frac{u}{v}\right), s = 2uv, r = \ln\left(\frac{v}{2u}\right)$$ Find: - Number of chains and terms in each chain. - Compute the chains in terms of variables. - Find $$\frac{\partial \mathbf{w}}{\partial u}$$ and $$\frac{\partial \mathbf{w}}{\partial v}$$ at $$u=1$$ and $$v=1+9=10$$ (last digit of roll number 49 is 9). 2. **Step 1: Number of chains and terms** - The function $$\mathbf{w}$$ depends on $$t, r, s$$. - Each of $$t, r, s$$ depends on $$u$$ and $$v$$. - So, there are 3 chains: $$t(u,v), r(u,v), s(u,v)$$. - Each chain has 2 terms (variables $$u$$ and $$v$$). 3. **Step 2: Compute chains in terms of variables** - $$t = \tan^{-1}\left(\frac{u}{v}\right)$$ - $$s = 2uv$$ - $$r = \ln\left(\frac{v}{2u}\right) = \ln(v) - \ln(2u)$$ 4. **Step 3: Find partial derivatives of $$\mathbf{w}$$ components w.r.t $$t, r, s$$** - $$\mathbf{w} = \begin{bmatrix}4t \\ r \\ s\end{bmatrix}$$ - $$\frac{\partial w_1}{\partial t} = 4, \quad \frac{\partial w_1}{\partial r} = 0, \quad \frac{\partial w_1}{\partial s} = 0$$ - $$\frac{\partial w_2}{\partial t} = 0, \quad \frac{\partial w_2}{\partial r} = 1, \quad \frac{\partial w_2}{\partial s} = 0$$ - $$\frac{\partial w_3}{\partial t} = 0, \quad \frac{\partial w_3}{\partial r} = 0, \quad \frac{\partial w_3}{\partial s} = 1$$ 5. **Step 4: Find partial derivatives of $$t, r, s$$ w.r.t $$u$$ and $$v$$** - For $$t = \tan^{-1}\left(\frac{u}{v}\right)$$: $$\frac{\partial t}{\partial u} = \frac{1}{1 + (\frac{u}{v})^2} \cdot \frac{1}{v} = \frac{v}{v^2 + u^2}$$ $$\frac{\partial t}{\partial v} = \frac{1}{1 + (\frac{u}{v})^2} \cdot \left(-\frac{u}{v^2}\right) = -\frac{u}{v^2 + u^2}$$ - For $$s = 2uv$$: $$\frac{\partial s}{\partial u} = 2v$$ $$\frac{\partial s}{\partial v} = 2u$$ - For $$r = \ln(v) - \ln(2u)$$: $$\frac{\partial r}{\partial u} = -\frac{1}{u}$$ $$\frac{\partial r}{\partial v} = \frac{1}{v}$$ 6. **Step 5: Use chain rule to find $$\frac{\partial \mathbf{w}}{\partial u}$$ and $$\frac{\partial \mathbf{w}}{\partial v}$$** - For $$\frac{\partial \mathbf{w}}{\partial u}$$: $$\frac{\partial w_1}{\partial u} = 4 \cdot \frac{\partial t}{\partial u} = 4 \cdot \frac{v}{v^2 + u^2}$$ $$\frac{\partial w_2}{\partial u} = 1 \cdot \frac{\partial r}{\partial u} = -\frac{1}{u}$$ $$\frac{\partial w_3}{\partial u} = 1 \cdot \frac{\partial s}{\partial u} = 2v$$ - For $$\frac{\partial \mathbf{w}}{\partial v}$$: $$\frac{\partial w_1}{\partial v} = 4 \cdot \frac{\partial t}{\partial v} = 4 \cdot \left(-\frac{u}{v^2 + u^2}\right) = -\frac{4u}{v^2 + u^2}$$ $$\frac{\partial w_2}{\partial v} = 1 \cdot \frac{\partial r}{\partial v} = \frac{1}{v}$$ $$\frac{\partial w_3}{\partial v} = 1 \cdot \frac{\partial s}{\partial v} = 2u$$ 7. **Step 6: Substitute $$u=1$$ and $$v=10$$** - Calculate denominators: $$v^2 + u^2 = 10^2 + 1^2 = 100 + 1 = 101$$ - Compute $$\frac{\partial \mathbf{w}}{\partial u}$$: $$\frac{\partial w_1}{\partial u} = 4 \cdot \frac{10}{101} = \frac{40}{101}$$ $$\frac{\partial w_2}{\partial u} = -\frac{1}{1} = -1$$ $$\frac{\partial w_3}{\partial u} = 2 \cdot 10 = 20$$ - Compute $$\frac{\partial \mathbf{w}}{\partial v}$$: $$\frac{\partial w_1}{\partial v} = -\frac{4 \cdot 1}{101} = -\frac{4}{101}$$ $$\frac{\partial w_2}{\partial v} = \frac{1}{10} = 0.1$$ $$\frac{\partial w_3}{\partial v} = 2 \cdot 1 = 2$$ **Final answers:** $$\frac{\partial \mathbf{w}}{\partial u} = \begin{bmatrix} \frac{40}{101} \\ -1 \\ 20 \end{bmatrix}, \quad \frac{\partial \mathbf{w}}{\partial v} = \begin{bmatrix} -\frac{4}{101} \\ 0.1 \\ 2 \end{bmatrix}$$