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Chapter 11: Problem 14

Find the directional derivative of the function at \(P\) in the direction of \(Q\). $$ h(x, y, z)=\ln (x+y+z), \quad P(1,0,0), Q(4,3,1) $$

Short Answer

Expert verified
The directional derivative of the function at P in the direction of Q is \(7/\sqrt{19}\).

Step by step solution

01

Find the Gradient of h

The gradient of h, denoted 鈭噃, is the vector containing the partial derivatives of h with respect to each variable: \n鈭噃 = (鈭俬/鈭倄, 鈭俬/鈭倅, 鈭俬/鈭倆). \nWhen applying the gradient to function h(x, y, z) = ln(x + y + z), we obtain 鈭噃 = (1/(x + y + z), 1/(x + y + z), 1/(x + y + z)). At point P(1, 0, 0), the gradient becomes 鈭噃(P) = (1/1, 1/1, 1/1) = (1, 1, 1).
02

Calculate the Unit Vector in the Direction of Q from P

The unit vector from point P to Q can be calculated as follows:\nDetermine the vector PQ = Q - P = (4 - 1, 3 - 0, 1 - 0) = (3, 3, 1).\nThen, the unit vector from P to Q, or u, is obtained by dividing PQ by its magnitude.\nThe magnitude of a vector (a, b, c) is given by sqrt(a虏 + b虏 + c虏).\nFor the vector PQ = (3, 3, 1), its magnitude is sqrt(3虏 + 3虏 + 1虏) = sqrt(19).\nSo, the unit vector u is (3/sqrt(19), 3/sqrt(19), 1/sqrt(19)).
03

Compute the Directional Derivative

The directional derivative of h at P in the direction of Q is given by 鈭噃(P) . u, which is the dot product of the gradient of h at P and the unit vector in the direction from P to Q.\nSo, 鈭噃(P) . u is (1, 1, 1) . (3/sqrt(19), 3/sqrt(19), 1/sqrt(19)) = 1*(3/sqrt(19)) + 1*(3/sqrt(19)) + 1*(1/sqrt(19)) = (7/sqrt(19)).

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Most popular questions from this chapter

Find \(d w / d t\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(t\) before differentiating. \(w=x y \cos z, \quad x=t, \quad y=t^{2}, \quad z=\arccos t\)

In Exercises 51-58, differentiate implicitly to find the first partial derivatives of \(z\) \(x^{2}+y^{2}+z^{2}=25\)

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