Question

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.)

Answer 1

The directional derivative of a function $f(x,y)$ in the direction of $\mathbf v$ is given by

$\nabla f(x,y)\cdot\mathbf v$

We have $\vec{ab}=\mathbf b-\mathbf a=(12-7,3-3)=(5,0)$, so that $\|\vec{ab}\|=5$, at which point we're given

$\nabla f(7,3)\cdot\dfrac{(5,0)}5=5\implies1\cdot\dfrac{\partial f}{\partial x}(7,3)+0\cdot\dfrac{\partial f}{\partial y}(7,3)=5$

$\implies\dfrac{\partial f}{\partial x}(7,3)=5$

We're also given that, in the direction of $\vec{ac}=\mathbf c-\mathbf a=(7-7,7-3)=(0,4)$ with $\|\vec{ac}\|=4$, we have

$\nabla f(7,3)\cdot\dfrac{(0,4)}4=4\implies0\cdot\dfrac{\partial f}{\partial x}(7,3)+1\cdot\dfrac{\partial f}{\partial y}(7,3)=4$

$\implies\dfrac{\partial f}{\partial y}(7,3)=4$

So in the direction of $\vec{ad}=\mathbf d-\mathbf a=(15-7,9-3)=(8,6)$, with $\|\vec{ad}\|=10$, we have

$\nabla f(7,3)\cdot\dfrac{(8,6)}{10}=\dfrac1{10}(4,4)\cdot(8,6)=9.60$

Answer 2

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

Related Link :https://brainly.com/question/7638557