Calculus: rate of change at a point when given a direction

label Calculus
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Given f(x, y) = 3x^2 + xy + y^2

Find the gradient ∇f(1, 1).

Then Find the rate of change of f(x, y) at the point (1, 1) in the direction of (1, 2)

Aug 5th, 2015


0. Let's find the partial derivatives of f.

(d/dx)f(x,y) = 6x + y,
(d/dy)f(x,y) = x + 2y.

So (d/dx)f(1,1) = 7, (d/dy)f(1,1) = 3.

1. Gradient is a vector formed from the partial derivatives. Therefore the answer is
<7, 3>

2. The rate of change is equal to a*(d/dx)f(x,y) + b*(d/dy)f(x,y)
where <a,b> is the direction vector of the length 1.

We have direction vector of the form <1, 2>. Its length is sqrt(5) so the corresponding unit vector (of the length 1) is <1/sqrt(5), 2/sqrt(5)>.

The rate of change at the point (1,1) is equal to
(1/sqrt(5))*7 + (2/sqrt(5))*3 = 13/sqrt(5) = 13*sqrt(5)/5 which is approx = 5.81.

Please ask if anything is unclear.
Aug 5th, 2015

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