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

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 (a)
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. (a) Find the velocity (vector) of the bee at time t = 2.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Write the velocity vector $\displaystyle v$ of the bee as $\displaystyle v(t)=[x'(t),y'(t),z'(t)]$ for the time-dependent space-functions $\displaystyle x(t),y(t),z(t)$ and use that $\displaystyle [x(t),y(t),z(t)]$ satisfies $\displaystyle 3z+x^{2}+y^{2}=2$ and $\displaystyle z=x^{2}-y^{2}$

Hint 2

The derivatives of $\displaystyle x(t),y(t),z(t)$ must satisfy $\displaystyle 3z'(t)+2x(t)x'(t)+2y'(t)y(t)=0$ and $\displaystyle 2x'(t)x(t)=z'(t)+2y(t)y'(t)$ Now you can plug in the point in question $\displaystyle (1,1,0)$ for $\displaystyle (x(t),y(t),z(t))$ and solve the two resulting equations for $\displaystyle x'(t)$ or $\displaystyle y'(t)$ or $\displaystyle z'(t)$. Write the velocity vector in terms of the derivative you chose (lets say $\displaystyle z'(t)$), and use $\displaystyle |v|=6$ to calculate the value of $\displaystyle z'(t)$.

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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. We let the vector $\mathbf {v}$ be the velocity vector. If the bee travels along a path in $\mathbb {R} ^{3}$ given by $[x(t),y(t),z(t)]^{T}$, then $\mathbf {v}$ is given by $\mathbf {v} ={\begin{bmatrix}{\frac {d}{dt}}x(t),{\frac {d}{dt}}y(t),{\frac {d}{dt}}z(t)\end{bmatrix}}^{T}$ We are told that the bee is travelling along a path defined by the intersection of the two surfaces: $3z+x^{2}+y^{2}=2$ and $z=x^{2}-y^{2}$. Therefore, we can differentiate each relationship with respect to t to get a pair of equations describing the velocity of the bee at any point along the curve: $3{\frac {dz}{dt}}+2x{\frac {dx}{dt}}+2y{\frac {dy}{dt}}=0,\quad {\frac {dz}{dt}}=2x{\frac {dx}{dt}}-2y{\frac {dy}{dt}}.$ We are interested in analyzing the velocity at (1,1,0), so plug this point into the above equations: $3{\frac {dz}{dt}}+2{\frac {dx}{dt}}+2{\frac {dy}{dt}}=0,\quad {\frac {dz}{dt}}-2{\frac {dx}{dt}}+2{\frac {dy}{dt}}=0.$ Solving the two equations to get ${\frac {dx}{dt}},\,{\frac {dy}{dt}}$ in terms of ${\frac {dz}{dt}}$ gives: ${\frac {dx}{dt}}=-{\frac {1}{2}}{\frac {dz}{dt}},\quad {\frac {dy}{dt}}=-{\frac {dz}{dt}}$ So at the point (1,1,0), we can write the velocity vector in terms of $z'(t)$: $\mathbf {v} =\left({\frac {d}{dt}}z(t)\right){\begin{bmatrix}-{\frac {1}{2}},-1,1\end{bmatrix}}^{T}.$ We know that the speed of the bee is 6 at the point (1,1,0), so we use the equation $|\mathbf {v} |=\left({\sqrt {\left({\frac {d}{dt}}z(t)\right)^{2}}}\right)\left({\sqrt {\left(-{\frac {1}{2}}\right)^{2}+\left(-1\right)^{2}+\left(1\right)^{2}}}\right)=6$ Solving for ${\frac {dz}{dt}}$ gives $\left({\sqrt {\left({\frac {d}{dt}}z(t)\right)^{2}}}\right)={\frac {6}{\sqrt {9/4}}}=4\quad \rightarrow \quad {\frac {d}{dt}}z(t)=\pm 4.$ Taking the positive solution, (since the bee is travelling in the direction of increasing z, we can write the velocity vector ${\color {blue}\mathbf {v} ={\begin{bmatrix}-2,-4,4\end{bmatrix}}^{T}}$

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