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- Consider the function $f\colon\R^2\to\R$ by $f(x,y)=x^2y^3$. Suppose you want to know the directional derivative for $f$ at the point $(2,-1)$ in the direction of $\langle 1,2\rangle$.
- In class, we found that the directional derivative in the direction of a unit vector $\u$ at $(x_0,y_0)$ is $\grad\!f\cdot\u$, where $\grad\!f = \langle f_x,f_y \rangle$, the vector of partial derivatives. Use this to compute the directional derivative indicated above.
(Hint: remember $\u$ must be a unit vector.)
- Last Friday, we described another method of determining the directional derivative at $(x_0,y_0)$ in the direction of a unit vector, $\u$: first parameterize the line through $(x_0,y_0)$ in the direction of $\u$ as a function $L(t)$, then plug this line into $f$ to get a new function $F\colon\R\to\R$ by $F(t)=f(L(t))$. Then since $L(0)=(x_0,y_0)$, $F'(0)$ will give the directional derivative at $(x_0,y_0)$ in the direction of $\u$. Use this approach to compute the directional derivative described above.
(Hint: again, you need to make sure $\u$ is a unit vector. Your answer should be the same as the one you obtained above. You may find it easier to use the product and chain rules rather than trying to multiply everything out before differentiating. Since you are going to be substituting $t=0$, there is not much point in simplifying the answer before evaluating it.)
- Which of these approaches is easier?
- In edition 10, Section 13.3 # 25, 27, 29, 31, 37, 55, 62
In edition 9, these are Section 13.3 # 17, 19, 21, 23, 29, 31, 47, 54
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