3 years ago

Point P is the midpoint of DE.

Mathematics

1 answer:

Leto [7]3 years ago

6 0

Answer:A.4

Step-by-step explanation:

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b. The gradient at point P(1, 2) is

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It looks like

$\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath$

so that

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Then

$D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}$

$\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}$

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