The value of zu(1,3) using the data elements represented on the table of values is q + p
<h3>How to solve the calculus expression?</h3>
The given parameters are:
z = f(x, y)
x = x(u, v)
y = y(u, v)
Where
x(1, 3) = 2 and y(1, 3) = 2
To calculate zu(1,3), we make use of:
![z_u(1,3) = f_x(x,y) * x_u(1,3) + f_y(x,y) * y_u(1,3)](https://tex.z-dn.net/?f=z_u%281%2C3%29%20%3D%20f_x%28x%2Cy%29%20%2A%20x_u%281%2C3%29%20%2B%20%20f_y%28x%2Cy%29%20%2A%20y_u%281%2C3%29)
The values x(1, 3) = 2 and y(1, 3) = 2 mean that:
(x,y) = (2,2).
So, we have:
![z_u(1,3) = f_x(2,2) * x_u(1,3) + f_y(2,2) * y_u(1,3)](https://tex.z-dn.net/?f=z_u%281%2C3%29%20%3D%20f_x%282%2C2%29%20%2A%20x_u%281%2C3%29%20%2B%20%20f_y%282%2C2%29%20%2A%20y_u%281%2C3%29)
From the table of values, we have:
![f_x(2,2) = q](https://tex.z-dn.net/?f=f_x%282%2C2%29%20%3D%20q)
![x_u(1,3) = 1](https://tex.z-dn.net/?f=x_u%281%2C3%29%20%3D%201)
![f_y(2,2) = 1](https://tex.z-dn.net/?f=f_y%282%2C2%29%20%3D%201)
![y_u(1,3) = p](https://tex.z-dn.net/?f=y_u%281%2C3%29%20%3D%20p)
So, the equation becomes
![z_u(1,3) = q * 1 + 1 * p](https://tex.z-dn.net/?f=z_u%281%2C3%29%20%3D%20q%20%2A%201%20%2B%20%201%20%2A%20p)
Evaluate the product
![z_u(1,3) = q + p](https://tex.z-dn.net/?f=z_u%281%2C3%29%20%3D%20q%20%2B%20p)
Hence, the value of zu(1,3) is q + p
Read more about calculus at:
brainly.com/question/5313449
#SPJ1