Find the value of constant(a+b+c)so that the directional derivative of the functionf=axy2+byz+cz2x3at the point (1, 2, –1) has maximum magnitude 64 in the direction parallel toyaxis :Correct answer is '-20'. Can you explain this answer?

Question

Find the value of constant(a+b+c)so that the directional derivative of the functionf=axy2+byz+cz2x3at the point (1, 2, &ndash;1) has maximum magnitude 64 in the direction parallel toyaxis :Correct answer is '-20'. Can you explain this answer?

Answer

To find the value of the constant (a, b, c) that would give the desired directional derivative, we need to use the formula for the directional derivative:

D_v(f) = ∇f · v

where D_v(f) is the directional derivative of the function f, ∇f is the gradient of f, and v is the direction vector.

First, let's find the gradient of f. The gradient of f is given by:

∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z)

Taking the partial derivatives of f with respect to x, y, and z, we get:

∂f/∂x = ay^2 z c z^2
∂f/∂y = ax z^2 c z^2
∂f/∂z = axy^2 cz^2

Now, let's find the direction vector v. The direction vector is given by the components (a, b, c).

Using the formula for the directional derivative, we have:

D_v(f) = ∇f · v
= (∂f/∂x, ∂f/∂y, ∂f/∂z) · (a, b, c)
= (ay^2 z c z^2, ax z^2 c z^2, axy^2 cz^2) · (a, b, c)
= a(ay^2 z c z^2) + b(ax z^2 c z^2) + c(axy^2 cz^2)
= a^2 y^2 z^3 c + abx z^4 c + acx y^2 z^4 c^2

To find the value of the constant (a, b, c), we need to set the directional derivative equal to the desired value. Let's say the desired directional derivative is D_v(f) = k.

k = a^2 y^2 z^3 c + abx z^4 c + acx y^2 z^4 c^2

Since we are given the point (1, 2, 3), let's substitute these values into the equation:

k = a^2 (2)^2 (3)^3 c + a(1)(3)^4 c + a(1)(2)^2 (3)^4 c^2
k = 36a^2 c + 81ac + 5832ac^2

Now, we can solve this equation for the constant (a, b, c) by setting it equal to the desired value of k and solving for (a, b, c).

Similar Physics Doubts

If the scalar function isf = x2 y z3than at the point (2, 1,-1)

Find the directional derivative of v^(2) at the point (2 0 3) in the direction of the outer normal of the sphere x^(2) y^(2) z^(2)=14 at the point (3 2 1) for v=xy^(2)y zy^(2j xz^(2)k.?

Find the directional derivative of the point (2, 0, 3) is the direction of the outward normal to the sphere x2 + y2 + z2 = 14 at the points (3, 4, 0) ? Correct answer is '540'. Can you explain this answer?

A Positive point charge is moving in clockwise direction in a circle with constant speed. Consider the magnetic field produced by the charge at a pointP(not centre of the circl e) on the axis of the circle. Which of the following is incorrect.

Top Courses for Physics

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams