Asked by Sara
Find the gradient and directional derivative of f at P in the direction of v:
f(x,y)=ln(x^2 + y^2 +1)+ e^(2xy)
P(0,2)
v=5i-12j
f(x,y)=ln(x^2 + y^2 +1)+ e^(2xy)
P(0,2)
v=5i-12j
Answers
Answered by
Steve
The gradient ∇f
= 2x/(x^2+y^2+1) + 2ye^(2xy)<b>i</b>
+ 2y/(x^2+y^2+1) + 2xe^(2xy)<b>j</b>
The directional derivative along <b>v</v> is ∇f•<b>v</v>
So, at (0,2)
Fx = 4
Fy = 4/5
and
∇f•<b>v</v> = <4,4/5>•<5,-12> = <20,-48/5></b></b></b>
= 2x/(x^2+y^2+1) + 2ye^(2xy)<b>i</b>
+ 2y/(x^2+y^2+1) + 2xe^(2xy)<b>j</b>
The directional derivative along <b>v</v> is ∇f•<b>v</v>
So, at (0,2)
Fx = 4
Fy = 4/5
and
∇f•<b>v</v> = <4,4/5>•<5,-12> = <20,-48/5></b></b></b>
Answered by
Steve
Sorry about the boldface.
The gradient ∇f
= 2x/(x^2+y^2+1) + 2ye^(2xy)<b>i</b>
+ 2y/(x^2+y^2+1) + 2xe^(2xy)<b>j</b>
The directional derivative along <b>v</v> is ∇f•<b>v</b>
So, at (0,2)
Fx = 4
Fy = 4/5
and
∇f•<b>v</b> = <4,4/5>•<5,-12> = <20,-48/5></b>
The gradient ∇f
= 2x/(x^2+y^2+1) + 2ye^(2xy)<b>i</b>
+ 2y/(x^2+y^2+1) + 2xe^(2xy)<b>j</b>
The directional derivative along <b>v</v> is ∇f•<b>v</b>
So, at (0,2)
Fx = 4
Fy = 4/5
and
∇f•<b>v</b> = <4,4/5>•<5,-12> = <20,-48/5></b>
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