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Find E(x,y) for the function x³+y³=3/2xyz at the point (1,2,3) . Round your answer coefficients to at least 4 decimal places.

User Haneulkim
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To find the gradient of the function at a given point, we first need to find the partial derivatives of the function with respect to x and y. Our function in this case is:

f(x, y, z) = x³ + y³ - (3/2)xyz

Let's differentiate the function first with respect to x, holding y and z constant. This gives us the x-component of our gradient.

Differentiating the first term gives us 3x². The second term (y³) is constant with respect to x and its derivative is 0. The third term is a product of x, y and z where z is a constant and y is treated as a constant. So we apply the product rule.

The derivative of the third term is therefore: (3/2)y*z - (3/2)*yz = (3/2)y*z - 3/2*x*y

Our full derivative of f with respect to x is then:

fₓ(x, y, z) = 3x² - (3/2)*yz

Let's evaluate fₓ at (1, 2, 3). Substituting these values in gives:

fₓ(1, 2, 3) = 3(1)² - (3/2)*(2)*(3) = 3 - 9 = -6

Now let's find the derivative with respect to y (f_y). Differentiating each term of f(x, y, z) we get:

f_y(x, y, z) = 0 + 3y² - (3/2)*xz

When we substitute the point (1, 2, 3) into this equation:

f_y(1, 2, 3) = 3(2)² - (3/2)*(1)*(3) = 12 - 4.5 = 7.5

So, the E(x, y) at point (1, 2, 3) is -6 and 7.5 which is (-6.0000, 7.5000) when rounded to 4 decimal places.

User Aabha Pandey
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