Let f be a function of two variables that has continuous partial derivatives and consider the points a(7, 3), b(12, 3), c(7, 7), and d(15, 9). the directional derivative of f at a in the direction of the vector ab is 5 and the directional derivative at a in the direction of ac is 4. find the directional derivative of f at a in the direction of the vector ad. (round your answer to two decimal places.)

To answer this question, it is necessary to make use of the concepts of directional derivative and gradient of a function.

The solution is:

grad F( x, y ) ×ad(u)  =64/10

The gradient of a function f (x,y)  is a vector defined by:

grad f(x,y) = δf(x,y)/δx × i + δf(x,y)/δy × j

and directional derivative, in the direction of ab is defined by :

grad f( x,y) × Uᵃᵇ(u)   (1)    where Uᵃᵇ is a unitary vector in the direction of ab

according to that    Uᵃᵇ (u) = Uᵃᵇ/|ab|

Then if the directional derivative of f (x,y) in the direction of the vector ab is 5.

A ( 7 , 3 )   B ( 12 , 3 )    then   vector ab is:

ab = [ 12 - 7 , 3 - 3 ]     ⇒ ab = [ 5 , 0 ]  and |ab| = √ (5)² (0)²   |ab| = 5

a unitary vector in the direction of ab is:

ab(u) = 5×i/ 5 and according to equation (1)

δf(x,y) / δx × i  × 5 × i / |5| = 5

δf(x,y) / δx × i  × 5 × i = 25

δf(x,y) / δx = 5       then   f( x,y ) = 5 × x  +

We go on to calculate the component on j of f(x,y)

Following the same procedure

ac = ( 7 , 7 ) - ( 7 , 3 )     ⇒   ac = [ 0 ,  4 ]      |ac| = √(4)² + (0)²

|ac| = 4

Unitary vector in the direction of ac(u)  is:

ac/|ac|  = 4 × j / 4

Then :

δf(x,y) / δy × j ×  + 4 × j / 4 = 4

δf(x,y) / δy × j ×  + 4 × j  = 16

δf(x,y) / δy = - 4       and     f(x,y) =  -4×y

f(x,y) = 5 × i + 4 × j

Finally:

vector ad = [ ( 15 - 7 , 9 - 3 ) ]     ⇒  ad = ( 8 , 6 )

Unitary vector in direction ad is

ad(u) = ( 8 ×i  + 6 ×j ) / √ (8)² + (6)²     ⇒  ad(u) = ( 8 ×i  + 6 ×j ) /√ (8)² + (6)²

ad(u)  = ( 8 ×i  + 6 ×j ) /10

Now we have f (x,y ) = 5 × i + 4 × j   and ad(u) = ( 8 ×i  + 6 ×j ) /10

We can calculate the directional derivative of f(x,y) in the direction of ad with the use of equation (1)

grad F( x, y ) ×ad(u)  = ( 5 × i + 4 × j ) × ( 8 ×i  + 6 ×j ) /10

grad F( x, y ) ×ad(u)  = 4 + ( 24/10)

grad F( x, y ) ×ad(u)  =64/10

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step-by-step explanation:

1) 2/5

2) zero=0/1

3) zero=0/2

4) -4/5

5) undefined=3/0