Question

Let \(T(x,y,z)=xy^2+2z-x^2z^2\) be the temperature at the point \((x,y,z)\). The unit vector in the direction in which the temperature decreases most rapidly at \((1,0,-1)\) is

• A. \(\dfrac{\hat i-4\hat k}{\sqrt{17}}\)
• B. \(\dfrac{\hat i-2\hat k}{\sqrt{5}}\)
• C. \(\dfrac{-\hat i+2\hat k}{\sqrt{5}}\)
• D. \(\dfrac{2\hat i+3\hat j+\hat k}{\sqrt{14}}\)

Hint:

Direction of maximum increase is \(\nabla T\), while direction of maximum decrease is \(-\nabla T\).

Answer 1

The Correct Option is B

Concept:
The temperature increases most rapidly in the direction of the gradient \[ \nabla T \] and decreases most rapidly in the opposite direction \[ -\nabla T \] The required answer is the unit vector along \[ -\nabla T \]

Step 1: Find the gradient of \(T\). Given, \[ T(x,y,z)=xy^2+2z-x^2z^2 \] \[ \nabla T= \frac{\partial T}{\partial x}\hat i+ \frac{\partial T}{\partial y}\hat j+ \frac{\partial T}{\partial z}\hat k \]

Step 2: Compute partial derivatives. \[ \frac{\partial T}{\partial x}=y^2-2xz^2 \] \[ \frac{\partial T}{\partial y}=2xy \] \[ \frac{\partial T}{\partial z}=2-2x^2z \]

Step 3: Substitute \((1,0,-1)\). At \[ x=1,\quad y=0,\quad z=-1 \] \[ \frac{\partial T}{\partial x}=0^2-2(1)(-1)^2 \] \[ =-2 \] \[ \frac{\partial T}{\partial y}=2(1)(0)=0 \] \[ \frac{\partial T}{\partial z}=2-2(1)^2(-1) \] \[ =2+2 \] \[ =4 \] Thus, \[ \nabla T=-2\hat i+4\hat k \]

Step 4: Direction of most rapid decrease. \[ -\nabla T=2\hat i-4\hat k \] This is proportional to \[ \hat i-2\hat k \]

Step 5: Convert into unit vector. Magnitude of \[ \hat i-2\hat k \] is \[ \sqrt{1^2+(-2)^2} \] \[ =\sqrt{5} \] Therefore, the required unit vector is \[ \frac{\hat i-2\hat k}{\sqrt5} \]

Step 6: Final answer. \[ \boxed{\frac{\hat i-2\hat k}{\sqrt5}} \]