Science:Math Exam Resources/Courses/MATH200/December 2011/Question 03 (b)

MATH200 December 2011

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q4  •  Q5 (a)  •  Q5 (b)  •  Q6  •  Q7  •  Q8 (a)  •  Q8 (b) i  •  Q8 (b) ii  •  Q8 (b) iii  •  

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  •  April 2012  •  December 2011  •  

Question 03 (b)
A bee is flying along the curve of intersection of the surfaces $3z+x^{2}+y^{2}=2$ and $z=x^{2}-y^{2}$ in the direction for which z is increasing. At time t = 2, the bee passes through the point (1,1,0) at speed 6. (b) The temperature T at position (x,y,z) at time t is given by $T=xy-3x+2yt+z$. Find the rate of change of temperature experienced by the bee at time $t=2$.

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Hint
The rate of change of the temperature is the derivative of the temperature, which is the partial derivative of the temperature plus the directional derivative of $\displaystyle T$ in the direction of the velocity vector v of the bee, ${\frac {dT}{dt}}={\frac {\partial T}{\partial t}}+\nabla T\cdot \mathbf {v}$ The vector v for the time $t=2$ is already determined in part (a).

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. To evaluate the change in temperature, $T$, experienced by the bee with respect to time at time 2, we need to evaluate the full derivative of $\displaystyle T=T(x(t),y(t),z(t),t)$ with respect to $t$ and substitute in the value $t=2$. Writing the expression for the full time derivative gives: ${\begin{aligned}{\frac {dT}{dt}}&={\frac {\partial T}{\partial t}}+{\frac {\partial T}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial T}{\partial y}}{\frac {dy}{dt}}+{\frac {\partial T}{\partial z}}{\frac {dz}{dt}}\\&={\frac {\partial T}{\partial t}}+\nabla T\cdot \mathbf {v} \end{aligned}}$ where $\mathbf {v}$ is the velocity vector of the bee. Evaluating the terms in the above equation we find: ${\frac {\partial T}{\partial t}}=2y,\quad \nabla T=\left[y-3,x+2t,1\right]^{T}.$ Note that we don't need to re-evaluate $\mathbf {v}$ since we can just use our results from part (a) here. At time $t=2$, the bee is at $[1,1,0]$, so the above terms satisfy, ${\frac {\partial T}{\partial t}}=2,\quad \nabla T=\left[-2,5,1\right]^{T}$ at $t=2$. Using this and the fact that $\mathbf {v}$ $=[-2,-4,4]$, we evaluate the change in temperature: ${\begin{aligned}{\frac {dT}{dt}}&=2+[-2,5,1]^{T}\cdot [-2,-4,4]^{T}=2+4-20+4=-10.\end{aligned}}$ Therefore, the bee is experiencing a change in temperature of -10 temperature units per second at t = 2.

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