1. **Stating the problem:** We have a vector function $$\mathbf{w} = \begin{pmatrix} 4t \\ r \\ s \end{pmatrix}$$ where $$t = \tan^{-1}\left(\frac{u}{v}\right),\quad s = 2uv,\quad r = \ln\left(\frac{v}{2u}\right).$$ We need to: - Determine the number of chains and terms in each chain. - Compute the respective chains in terms of variables. - Find the partial derivatives $$\frac{\partial \mathbf{w}}{\partial u}$$ and $$\frac{\partial \mathbf{w}}{\partial v}$$ at $$u=1$$ and $$v=1+9=10$$ (since last digit of registration number is 9). 2. **Number of chains and terms:** Each component of $$\mathbf{w}$$ depends on intermediate variables $$t, r, s$$ which themselves depend on $$u$$ and $$v$$. - The chains are from $$u,v \to t,r,s \to w_i$$. - There are 3 chains corresponding to $$w_1=4t$$, $$w_2=r$$, and $$w_3=s$$. - Each chain has 2 terms (variables $$u$$ and $$v$$) feeding into $$t,r,s$$. 3. **Compute the chains:** - For $$t = \tan^{-1}\left(\frac{u}{v}\right)$$: $$\frac{\partial t}{\partial u} = \frac{1}{1+\left(\frac{u}{v}\right)^2} \cdot \frac{1}{v} = \frac{v}{v^2+u^2}$$ $$\frac{\partial t}{\partial v} = \frac{1}{1+\left(\frac{u}{v}\right)^2} \cdot \left(-\frac{u}{v^2}\right) = -\frac{u}{v^2+u^2}$$ - For $$s = 2uv$$: $$\frac{\partial s}{\partial u} = 2v$$ $$\frac{\partial s}{\partial v} = 2u$$ - For $$r = \ln\left(\frac{v}{2u}\right) = \ln(v) - \ln(2u)$$: $$\frac{\partial r}{\partial u} = -\frac{1}{u}$$ $$\frac{\partial r}{\partial v} = \frac{1}{v}$$ 4. **Find $$\frac{\partial \mathbf{w}}{\partial u}$$ and $$\frac{\partial \mathbf{w}}{\partial v}$$:** Recall $$\mathbf{w} = \begin{pmatrix}4t \\ r \\ s \end{pmatrix}$$, so $$\frac{\partial \mathbf{w}}{\partial u} = \begin{pmatrix}4 \frac{\partial t}{\partial u} \\ \frac{\partial r}{\partial u} \\ \frac{\partial s}{\partial u} \end{pmatrix}, \quad \frac{\partial \mathbf{w}}{\partial v} = \begin{pmatrix}4 \frac{\partial t}{\partial v} \\ \frac{\partial r}{\partial v} \\ \frac{\partial s}{\partial v} \end{pmatrix}$$ Substitute $$u=1$$ and $$v=10$$: - Compute $$\frac{\partial t}{\partial u}$$: $$\frac{10}{10^2 + 1^2} = \frac{10}{100 + 1} = \frac{10}{101}$$ - Compute $$\frac{\partial t}{\partial v}$$: $$-\frac{1}{101}$$ - Compute $$\frac{\partial r}{\partial u} = -\frac{1}{1} = -1$$ - Compute $$\frac{\partial r}{\partial v} = \frac{1}{10} = 0.1$$ - Compute $$\frac{\partial s}{\partial u} = 2 \times 10 = 20$$ - Compute $$\frac{\partial s}{\partial v} = 2 \times 1 = 2$$ Therefore, $$\frac{\partial \mathbf{w}}{\partial u} = \begin{pmatrix}4 \times \frac{10}{101} \\ -1 \\ 20 \end{pmatrix} = \begin{pmatrix} \frac{40}{101} \\ -1 \\ 20 \end{pmatrix}$$ $$\frac{\partial \mathbf{w}}{\partial v} = \begin{pmatrix}4 \times -\frac{1}{101} \\ 0.1 \\ 2 \end{pmatrix} = \begin{pmatrix} -\frac{4}{101} \\ 0.1 \\ 2 \end{pmatrix}$$ **Final answers:** $$\boxed{\frac{\partial \mathbf{w}}{\partial u} = \begin{pmatrix} \frac{40}{101} \\ -1 \\ 20 \end{pmatrix}, \quad \frac{\partial \mathbf{w}}{\partial v} = \begin{pmatrix} -\frac{4}{101} \\ 0.1 \\ 2 \end{pmatrix}}$$