Question

Let z = sqrt(3x+4y). Then:

The rate of change in z at (3,5) as we change x but hold y fixed is , and
The rate of change in z at (3,5) as we change y but hold x fixed is ?

Answer 1

The rate of change in z at (3,5) as we change x but hold y fixed is dz/dx = (3/(2*sqrt(29))). The rate of change in z at (3,5) as we change y but hold x fixed is dz/dy = (4/(2*sqrt(29))).

Step-by-step explanation:

The rate of change in z at (3,5) as we change x but hold y fixed can be found by taking the derivative of z with respect to x. Let's differentiate z with respect to x:

dz/dx = (3/(2*sqrt(3x+4y))

Substituting the values of (3,5) into the derivative, we get:

dz/dx = (3/(2*sqrt(3*3 + 4*5)) = (3/(2*sqrt(9 + 20)) = (3/(2*sqrt(29)))

So the rate of change in z at (3,5) as we change x but hold y fixed is dz/dx = (3/(2*sqrt(29))).

The rate of change in z at (3,5) as we change y but hold x fixed can be found by taking the derivative of z with respect to y. Let's differentiate z with respect to y:

dz/dy = (4/(2*sqrt(3x+4y))

Substituting the values of (3,5) into the derivative, we get:

dz/dy = (4/(2*sqrt(3*3 + 4*5)) = (4/(2*sqrt(9 + 20)) = (4/(2*sqrt(29)))

So the rate of change in z at (3,5) as we change y but hold x fixed is dz/dy = (4/(2*sqrt(29))).

Answer 2

The rate of change in z at (3,5) as we change x but hold y fixed is (3/2)z, and the rate of change in z at (3,5) as we change y but hold x fixed is 2z.

Step-by-step explanation:

To find the rate of change in z at (3,5) as we change x but hold y fixed, we can differentiate the equation with respect to x.

Since z = √(3x+4y), the rate of change is given by:

dz/dx = (3/2)√(3x+4y) = (3/2)z

To find the rate of change in z at (3,5) as we change y but hold x fixed, we can differentiate the equation with respect to y.

Since z = √(3x+4y), the rate of change is given by:

dz/dy = 2√(3x+4y) = 2z