Let f be a function of two variables that has continuous partial derivatives and consider the pointsA(8, 9),

B(10, 9),

C(8, 10),

and

D(11, 13).

The directional derivative of f at A in the direction of the vector AB is 9 and the directional derivative at A in the direction of

AC is 2. Find the directional derivative of f at A in the direction of the vector AD.

(Round your answer to two decimal places.)

Answers

Answer 1
Answer:

Answer:

The directional derivative of f at A in the direction of \vec{u} AD is 7.

Step-by-step explanation:

Step 1:

Directional of a function f in direction of the unit vector \vec{u}=(a,b) is denoted by D\vec{u}f(x,y),

D\vec{u}f(x,y)=f_(x)\left ( x ,y\right ).a+f_(y)(x,y).b.

Now the given points are

A(8,9),B(10,9),C(8,10) and D(11,13),

Step 2:

The vectors are given as

AB = (10-8, 9-9),the direction is

\vec{u}_(AB) = (AB)/(\left \| AB \right \|)=(1,0)

AC=(8-8,10-9), the direction is

\vec{u}_(AC) = (AC)/(\left \| AC \right \|)=(0,1)

AC=(11-8,13-9), the direction is

\vec{u}_(AD) = (AD)/(\left \| AD \right \|)=\left ((3)/(5),(4)/(5)  \right )

Step 3:

The given directional derivative of f at A \vec{u}_(AB) is 9,

D\vec{u}_(AB)f=f_(x) \cdot 1 + f_(y)\cdot 0\nf_(x) =9

The given directional derivative of f at A \vec{u}_(AC) is 2,

D\vec{u}_(AB)f=f_(x) \cdot 0 + f_(y)\cdot 1\nf_(y) =2

The given directional derivative of f at A \vec{u}_(AD) is

D\vec{u}_(AD)f=f_(x) \cdot (3)/(5) + f_(y)\cdot (4)/(5)

D\vec{u}_(AD)f=9 \cdot (3)/(5) + 2\cdot (4)/(5)

D\vec{u}_(AD)f= (27+8)/(5) =7

The directional derivative of f at A in the direction of  \vec{u}_(AD) is  7.


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Step-by-step explanation:

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Answers

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Answers

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Step-by-step explanation:

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Answers

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Answer:

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