Science:Math Exam Resources/Courses/MATH200/December 2012/Question 05

MATH200 December 2012

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Question 05

The directional derivative of a function $\displaystyle w=f(x,y,z)$ at a point P in the direction of the vector ${\vec {i}}$ is $\displaystyle 2$, in the direction of the vector ${\vec {i}}+{\vec {j}}$ is $-{\sqrt {2}}$, and in the direction of the vector ${\vec {i}}+{\vec {j}}+{\vec {k}}$ is $-{\frac {5}{\sqrt {3}}}$. Find the direction in which the function $\displaystyle w=f(x,y,z)$ has the maximum rate of change at the point P. What is the maximum rate of change?

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

Remember, that $\displaystyle {\vec {i}}={\begin{pmatrix}1\\0\\0\end{pmatrix}}$, $\displaystyle {\vec {j}}={\begin{pmatrix}0\\1\\0\end{pmatrix}}$, $\displaystyle {\vec {k}}={\begin{pmatrix}0\\0\\1\end{pmatrix}}$. Further, remember that the directional derivative is of a function $\displaystyle f$ in direction $\displaystyle v$ at the point $\displaystyle p$ is $\displaystyle \ \ D_{v}f_{p}=\nabla f\cdot v{\frac {1}{||v||}}$

Hint 2
Using the expressions from the first hint, make three equations for the gradient of the function $\displaystyle f$, where you write the gradient as $\displaystyle \nabla f={\begin{pmatrix}a\\b\\c\end{pmatrix}}$ Use the three equations to find $\displaystyle a,b,c$.

Hint 3
The maximal rate of change is in the direction of the gradient vector. The maximal rate of change is $\displaystyle ||\nabla f(p)||$

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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 have the following three statements given $\displaystyle D_{\vec {i}}f_{p}=2,\ \ D_{{\vec {i}}+{\vec {j}}}f_{p}=-{\sqrt {2}},\ \ D_{{\vec {i}}+{\vec {j}}+{\vec {k}}}f_{p}=-{\frac {5}{\sqrt {3}}}$ The directional derivative is calculated through $\displaystyle D_{v}f_{p}=\nabla f\cdot {\frac {v}{||v||}}$ where $\displaystyle v$ is the direction vector. Now we express the gradient of $\displaystyle f$ in the points $\displaystyle p$ as $\displaystyle \nabla f(p)={\begin{pmatrix}a\\b\\c\end{pmatrix}}$ and obtain $\displaystyle D_{\vec {i}}f_{p}=\nabla f(p)\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}{\frac {1}{\sqrt {1^{2}+0^{2}+0^{2}}}}={\begin{pmatrix}a\\b\\c\end{pmatrix}}\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}=a=2$ $\displaystyle D_{{\vec {i}}+{\vec {j}}}f_{p}={\begin{pmatrix}2\\b\\c\end{pmatrix}}\cdot {\begin{pmatrix}1\\1\\0\end{pmatrix}}{\frac {1}{\sqrt {2}}}=(2+b){\frac {1}{\sqrt {2}}}=-{\sqrt {2}}\quad \Rightarrow \quad 2+b=-2\quad \Rightarrow \quad b=-4$ $\displaystyle D_{{\vec {i}}+{\vec {j}}+{\vec {k}}}f_{p}={\begin{pmatrix}2\\-4\\c\end{pmatrix}}\cdot {\begin{pmatrix}1\\1\\1\end{pmatrix}}{\frac {1}{\sqrt {3}}}=(2-4+c){\frac {1}{\sqrt {3}}}=-{\frac {5}{\sqrt {3}}}\quad \Rightarrow \quad 2-4+c=-5\quad \Rightarrow \quad c=-3$ So, we get that $\displaystyle \nabla f(p)={\begin{pmatrix}2\\-4\\-3\end{pmatrix}}$. The function $\displaystyle f$ increases most rapidly in the direction of the gradient vector. Hence, the required vector is $\displaystyle {\begin{pmatrix}2\\-4\\-3\end{pmatrix}}$. The value of the maximal directional derivative is $\displaystyle ||\nabla f(p)||={\sqrt {2^{2}+(-4)^{2}+(-3)^{2}}}={\sqrt {29}}$.

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