1. **Stating the problem:** We have a vector function $$\mathbf{w} = \begin{pmatrix}4t r s \\ 2r \\ r^2 + s^2 \end{pmatrix}$$ where $$t = \tan^{-1}\left(\frac{u}{v}\right), s = 2uv, r = \ln\left(\frac{v}{2u}\right).$$ We need to: - Determine the number of chains and terms in each chain. - Compute the 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 $$v=1 + \text{last digit of registration number}$$. 2. **Number of chains and terms:** - The function $$\mathbf{w}$$ depends on $$t, r, s$$ which themselves depend on $$u, v$$. - So, the chains are: $$u \to t, r, s \to \mathbf{w}$$ and $$v \to t, r, s \to \mathbf{w}$$. - There are 2 chains (one for $$u$$ and one for $$v$$). - Each chain has 3 terms: $$t, r, s$$. 3. **Compute the chains (partial derivatives of intermediate variables):** - $$t = \tan^{-1}\left(\frac{u}{v}\right)$$ $$\frac{\partial t}{\partial u} = \frac{v}{u^2 + v^2}$$ $$\frac{\partial t}{\partial v} = \frac{-u}{u^2 + v^2}$$ - $$s = 2uv$$ $$\frac{\partial s}{\partial u} = 2v$$ $$\frac{\partial s}{\partial v} = 2u$$ - $$r = \ln\left(\frac{v}{2u}\right) = \ln(v) - \ln(2u)$$ $$\frac{\partial r}{\partial u} = \frac{\partial}{\partial u}[-\ln(2u)] = -\frac{1}{u}$$ $$\frac{\partial r}{\partial v} = \frac{1}{v}$$ 4. **Compute partial derivatives of $$\mathbf{w}$$ components:** - $$w_1 = 4 t r s$$ Using product rule: $$\frac{\partial w_1}{\partial u} = 4 \left( \frac{\partial t}{\partial u} r s + t \frac{\partial r}{\partial u} s + t r \frac{\partial s}{\partial u} \right)$$ $$\frac{\partial w_1}{\partial v} = 4 \left( \frac{\partial t}{\partial v} r s + t \frac{\partial r}{\partial v} s + t r \frac{\partial s}{\partial v} \right)$$ - $$w_2 = 2 r$$ $$\frac{\partial w_2}{\partial u} = 2 \frac{\partial r}{\partial u} = 2 \left(-\frac{1}{u}\right) = -\frac{2}{u}$$ $$\frac{\partial w_2}{\partial v} = 2 \frac{\partial r}{\partial v} = \frac{2}{v}$$ - $$w_3 = r^2 + s^2$$ $$\frac{\partial w_3}{\partial u} = 2 r \frac{\partial r}{\partial u} + 2 s \frac{\partial s}{\partial u} = 2 r \left(-\frac{1}{u}\right) + 2 s (2 v)$$ $$\frac{\partial w_3}{\partial v} = 2 r \frac{\partial r}{\partial v} + 2 s \frac{\partial s}{\partial v} = 2 r \frac{1}{v} + 2 s (2 u)$$ 5. **Evaluate at $$u=1$$ and $$v=1 + d$$ where $$d$$ is the last digit of registration number. For example, if $$d=3$$, then $$v=4$$. Calculate intermediate values: - $$t = \tan^{-1}\left(\frac{1}{1+d}\right)$$ - $$s = 2 \times 1 \times (1+d) = 2(1+d)$$ - $$r = \ln\left(\frac{1+d}{2 \times 1}\right) = \ln\left(\frac{1+d}{2}\right)$$ 6. **Substitute and compute:** - $$\frac{\partial t}{\partial u} = \frac{1+d}{1 + (1+d)^2}$$ - $$\frac{\partial t}{\partial v} = \frac{-1}{1 + (1+d)^2}$$ - $$\frac{\partial s}{\partial u} = 2(1+d)$$ - $$\frac{\partial s}{\partial v} = 2$$ - $$\frac{\partial r}{\partial u} = -1$$ - $$\frac{\partial r}{\partial v} = \frac{1}{1+d}$$ 7. **Final partial derivatives:** - $$\frac{\partial w_1}{\partial u} = 4 \left( \frac{1+d}{1+(1+d)^2} \cdot r \cdot s + t \cdot (-1) \cdot s + t \cdot r \cdot 2(1+d) \right)$$ - $$\frac{\partial w_1}{\partial v} = 4 \left( \frac{-1}{1+(1+d)^2} \cdot r \cdot s + t \cdot \frac{1}{1+d} \cdot s + t \cdot r \cdot 2 \right)$$ - $$\frac{\partial w_2}{\partial u} = -2$$ - $$\frac{\partial w_2}{\partial v} = \frac{2}{1+d}$$ - $$\frac{\partial w_3}{\partial u} = 2 r (-1) + 2 s (2(1+d)) = -2 r + 4 s (1+d)$$ - $$\frac{\partial w_3}{\partial v} = 2 r \frac{1}{1+d} + 2 s (2) = \frac{2 r}{1+d} + 4 s$$ This completes the solution for the first question.