Question

In Exercises 19 and \(20,\) use the gradient to find the directional derivative of the function at \(P\) in the direction of \(Q\). $$ g(x, y)=x^{2}+y^{2}+1, \quad P(1,2), Q(3,6) $$

Short answer

The directional derivative of the function \(g(x, y) = x^2 + y^2 + 1\) at point \(P(1,2)\) in the direction of \(Q(3,6)\) is \(2\sqrt{5}\).

Step-by-step solution

Calculate the Gradient of the Function

The gradient of a function, often denoted \(\nabla f\) or grad \(f\), is a vector that contains all of the partial derivative information of the function. The gradient of the given function \(g(x, y) = x^2 + y^2 + 1\) is \(\[\nabla g = (2x, 2y)\]\). Substituting the coordinates of point P into the gradient, we get, \(\[\nabla g(1,2) = (2*1, 2*2) = (2, 4)\]\). So, the gradient of g at point P is the vector (2,4).

Find the Unit Vector for the Direction Q

In order to compute the directional derivative, we need a unit vector in the direction of Q. The unit vector is obtained by dividing each component of the vector by its magnitude. Here, the direction is determined by the vector formed from P to Q, which is \(\[Q - P = (3-1, 6-2) = (2, 4)\]\). The magnitude of this vector can be obtained as \(\[\sqrt{(2)^2 + (4)^2} = \sqrt{20}\]\). Therefore, the unit vector is \(\[\frac{(2,4)}{\sqrt{20}} = \(\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}})\]\).

Calculate the Directional Derivative

The directional derivative in the direction of vector Q is given by the dot product of the gradient vector and the unit vector of Q. Therefore, the directional derivative \(D_Q g(P)\) is as such: \(\[D_Q g(1,2) = \nabla g(1,2) \cdot unit\_vector\_Q = (2, 4) \cdot (\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}) = 2*(\frac{1}{\sqrt{5}}) + 4*(\frac{2}{\sqrt{5}}) = \frac{10}{\sqrt{5}} = 2\sqrt{5}\]\).