Topic: Heat Transfer
The temperature field in a body varies according to the equation $T(x,y)=x^{3}+4xy$. The direction of fastest variation in temperature at the point (1,0) is given by
- A. $3\hat{i}+8\hat{j}$
- B. $\hat{i}$
- C. $0.6\hat{i}+0.8\hat{j}$
- D. $0.5\hat{i}+0.866\hat{j}$
Answer: C. $0.6\hat{i}+0.8\hat{j}$
Explanation: For fastest variation, we calculate the gradient: $\nabla T = \frac{\partial T}{\partial x}\hat{i}+\frac{\partial T}{\partial y}\hat{j}$. $T=x^{3}+4xy$. $\frac{\partial T}{\partial x}=3x^{2}+4y$ and $\frac{\partial T}{\partial y}=4x$. At point (1,0): $(\frac{\partial T}{\partial x})_{(1,0)}=3$, $(\frac{\partial T}{\partial y})_{(1,0)}=4$. Gradient $=3\hat{i}+4\hat{j}$. For directional unit vector: $\hat{n}=\frac{3\hat{i}+4\hat{j}}{\sqrt{3^{2}+4^{2}}}=0.6\hat{i}+0.8\hat{j}$.