Logic will get you from A to B. Imagination will take you everywhere, Albert Einstein.
A function of two variables f: ℝ x ℝ → ℝ assigns to each ordered pair in its domain a unique real number, e.g., Area = $\frac{1}{2}b·h$, z = f(x, y) = 2x + 3y, f(x, y) = x^{2} + y^{2}, e^{x+y}, etc.
Partial derivatives are derivatives of a function of multiple variables, say f(x_{1}, x_{2}, ···, x_{n}) with respect to one of those variables, with the other variables held constant. They measure how the function changes as one variable changes, while keeping the others constant. $\frac{\partial f}{\partial x}(x_0, y_0) = \lim_{\Delta x \to 0} \frac{f(x_0+\Delta x, y_0)-f(x_0, y_0)}{\Delta x}$
Geometrically, the partial derivative $\frac{\partial f}{\partial x}(x_0, y_0)$ at a point (x_{0}, y_{0}) can be interpreted as the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane y = y_{0}. Similarly, $\frac{\partial f}{\partial y}(x_0, y_0)$ represents the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane x = x_{0}
$\frac{dw}{ds}\bigg|_{\vec{u}} = ∇w·\vec{u} = |∇w|·|\vec{u}|cos(θ) = |∇w|·cos(θ)$ where θ is the angle between the gradient and the given unit vector.
The gradient is the direction in which the function increases fastest (the direction of the steepest ascent) at a given point, and |∇w| = $\frac{dw}{ds}\bigg|_{\vec{u}=dir(∇w)}$.
The directional derivative is minimized when cos(θ) = -1 ↭ θ = 180 ↭ $\vec{u}$ is in the opposite direction of the gradient ∇w. Futhermore, $\frac{dw}{ds}\bigg|_{\vec{u}} = 0$ ↭ cos(θ) = 0 ↭ θ = 90° ↭ $\vec{u}$ ⊥ ∇w.
In multivariable calculus, we often encounter functions that depend on multiple variables. Sometimes, these variables are not independent of each other. Instead, they are related by certain equations or constraints. .
Consider a multivariable function f(x, y, z) subject to a constraint given by g(x, y, z) = c where c is a constant. This constraint allows us to express or solve for one of the variables in term of the others to understand their relationships.
Let’s look at a specific example to understand this concept better.
Consider the function f(x, y, z) = x^{2} + y^{2} + z^{2} subject to the constraint g(x, y, z) = x + y + z = 1. We can solve the constraint equation for one of the variables, namely z, to express it in terms of the others: z = z(x, y) = 1 -x -y.
By substituting this expression for z into the function f, we can reduce f to a function of two variables, x and y: f(x, y) = x^{2} + y^{2} + (1-x-y)^{2} =[Let’s expand and simplify this function] f(x, y) = x^{2} + y^{2} + 1 -2x -2y + x^{2} + 2xy + y^{2} = 2x^{2} + 2y^{2} + 2xy -2x -2y +1.
Understanding the relationship between the variables involves computing partial derivatives. The partial derivatives of z with respect to x and y are: $\frac{∂z}{∂x} = 4x + 2y -2, \frac{∂z}{∂y} = 4y +2x -2$.
To find the critical points, we set the partial derivatives to zero and solve the resulting equations:
$\begin{cases} 4x + 2y -2 = 0 (i) \\ 4y +2x -2 = 0(ii) \end{cases}$
From the first equation (i), 2x + y = 1. From the second equation (i), 2x + 4y = 2. Subtracting equation (1) from equation (2): 2x + 4y -(2x +y) = 2 -1 ↭ 3y = 1 ↭ $y = \frac{1}{3}$ ⇒[Substituting $y = \frac{1}{3}$ back into equation 1] $2x + \frac{1}{3} = 1 ⇒ 2x = \frac{2}{3}⇒ x = \frac{1}{3} ⇒ z = 1 -x -y = 1 -\frac{1}{3}-\frac{1}{3} = \frac{1}{3}$.
So the critical point is $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$.
When dealing with multivariable functions, it’s important to understand how small changes in each variable affect the function. This can be done by computing the total differential. Let’s explore this concept using the function g(x, y, z) = x^{2} + yz + z^{3} evaluated at the point (2, 3, 1).
g(2, 3, 1) = 2^{2} +3·1 + 1^{2} = 4 + 3 + 1 = 8.
To understand how small changes in x, y, z affect g, we compute the total differential dg.
The total differential dg of a multivariable function g(x, y, z) is given by dg = $\frac{∂g}{∂y}dx + \frac{∂g}{∂y}dy + \frac{∂g}{∂z}dz$.
We compute the partial derivatives of g: $\frac{∂g}{∂x} = 2x, \frac{∂g}{∂y} = z, \frac{∂g}{∂z} = y + 3z^2$
When differentiating with respect to a particular variable, we treat all other variables as constants.
Substituting the partial derivatives: $dg = 2xdx + zdy + (y+3z^2)dz$.
Next, we evaluate these partial derivatives at the point (2, 3, 1):
$\frac{∂g}{∂x}\bigg|_{(2, 3, 1)} = 2·2 = 4$
$\frac{∂g}{∂y}\bigg|_{(2, 3, 1)} = 1$
$\frac{∂g}{∂z}\bigg|_{(2, 3, 1)} = 3 + 3·1^2 = 6$
So, the total differential at (2, 3, 1) is: dg = 4dx +dy +6dz. Given g(x, y, z) is a constant (8), the total differential dg is zero: 4dx + dy + 6dz = 0 ⇒[Solving for dz] $dz = \frac{-1}{6}(4dx +dy) = \frac{-2}{3}dx +\frac{-1}{6}dy$.
If we interpret z as a function of x and y, z = z(x, y), then: dz = $\frac{∂z}{∂x}dx + \frac{∂z}{∂y}dz$. By comparing coefficients: $\frac{∂z}{∂x} = \frac{-2}{3}, \frac{∂z}{∂y} = \frac{-1}{6}$. This analysis shows how small changes in x and y affect z when the function g(x, y, z)= x^{2} +yz + z^{3} is held constant at 8.
When dealing with multivariable functions, it’s important to understand how to take derivatives with respect to one variable while treating other variables as constants. This concept can sometimes be confusing and the notation can become ambiguous, so let’s break it down.
$\frac{∂z}{∂x} = $[y is constant, hence dy = 0, $dz = \frac{-2}{3}dx +\frac{-1}{6}dy = \frac{-2}{3}dx$] $\frac{-2}{3}$. To avoid confusion, we can use explicit notation that indicates which variables are held constant.
For a general constraint g(x, y, z) = c, the total differential is $dg = \frac{∂g}{∂x}dx + \frac{∂g}{∂y}dy + \frac{∂g}{∂z}dz =[\text{Using a different notation, we can write}] g_xdx + g_ydy + g_zdz = 0$ ⇒[Solving for dz] $dz = -\frac{g_x}{g_z}dx -\frac{g_y}{g_z}dy$.
As we have previously reasoned, $\frac{∂z}{∂x}$ = [Set y = constant which implies dy = 0, then dz = $-\frac{g_x}{g_z}dx -\frac{g_y}{g_z}dy = -\frac{g_x}{g_z}dx$] $-\frac{g_x}{g_z}~\text{. Analogously,} \frac{∂z}{∂y} = -\frac{g_y}{g_z}$.
Consider the function f(x, y) = x + y. The partial derivative with respect to x is: $\frac{∂f}{∂x} =1$.
Now, if we change variables such as x = u, y = u + v, the function becomes f(x, y) = x + y = 2u +v. We compute the partial derivatives: $\frac{∂f}{∂x} =1, \frac{∂f}{∂u} = 2$.
Even though x = u, the partial derivatives differ because the condition of holding variables constant differ. When we take the partial derivate of a function f(x, y) with respect to x, $\frac{∂f}{∂x} = 1$, we are asking how f changes as we vary x while holding y constant. Similarly, when we take the partial derivate of a function f(u, v) with respect to u, $\frac{∂f}{∂u} = 2$ we are taking the partial derivative with respect to u while holding v constant.
Therefore, these notation can become confusing because they don’t make explicit what variable are being hold constant, we could clarify our previous notation, $(\frac{∂f}{∂x})_y$ that explicitly states that the partial derivative is taken with respect to x while holding y constant.
Assume f = f(x, y, z) where g(x, y, z) = c, $(\frac{∂f}{∂x})_y$ is the rate of change of f with respect to x, y = constant, x varies, and z = z(x, y). It provides clarity and eliminates ambiguity, but it is not especially pleasing to the eye 😄; and $(\frac{∂f}{∂u})_v$ states that the partial derivative is taken with respect to u while holding v constant.
Using this notation, it is clearly obvious that $(\frac{∂f}{∂x})_y = 1 ≠ (\frac{∂f}{∂u})_v = 2$
The goal is to calculate $(\frac{∂w}{∂y})_z$, which represents the partial derivative of w with respect to y while keeping z constant.
By the chain rule, the partial derivative of w with respect to y while keeping z constant can be expressed as: $(\frac{∂w}{∂y})_z = \frac{∂w}{∂x}\frac{∂x}{∂y} + \frac{∂w}{∂y}\frac{∂y}{∂y} + \frac{∂w}{∂z}\frac{∂z}{∂y}$.
We use the chain rule under the constraint that z is constant. This is crucial because z depends on both x and y: $\frac{∂z}{∂y} = 0,\text{and}\frac{∂y}{∂y} = 1$.
Thus, the expression simplifies to: $\frac{∂w}{∂y} = \frac{∂w}{∂x}\frac{∂x}{∂y} + \frac{∂w}{∂y}.$
Besides, from our original equation (i), let’s compute partial derivatives of w: $\frac{∂w}{∂x} = 4x, \frac{∂w}{∂y} = 6y$.
We use the constraint given in Equation (ii): $\frac{∂z}{∂y} =[\text{By the Chain Rule}] 2x(\frac{∂x}{∂y})+4y(\frac{∂y}{∂y}) = 2x(\frac{∂x}{∂y})+4y =$[z is constant, $\frac{∂z}{∂y} = 0$] 0 ⇒ $\frac{∂x}{∂y} = \frac{-4y}{2x} = \frac{-2y}{x}$.
Finally, combining all these intermediate calculations, we get:
$(\frac{∂w}{∂y})_z = \frac{∂w}{∂x}\frac{∂x}{∂y} + \frac{∂w}{∂y} = (4x)(\frac{-2y}{x})+6y = -8y + 6y = -2y.$ The partial derivative of w with respect to y, keeping z constant, is -2y.
The goal is to calculate $(\frac{∂w}{∂t})_{x, z}$, which represents the partial derivative of w with respect to t, while keeping x and z constant.
By the chain rule, the partial derivative of w with respect to t while keeping x and z constant can be expressed as: $(\frac{∂w}{∂t})_{x, z} = \frac{∂w}{∂x}\frac{∂x}{∂t} + \frac{∂w}{∂y}\frac{∂y}{∂t} + \frac{∂w}{∂z}\frac{∂z}{∂t} + \frac{∂w}{∂t}\frac{∂t}{∂t}$.
We use the chain rule under the constraint that x and z are constant, their partial derivatives with respect to t are zero: $\frac{∂x}{∂t} = 0,\frac{∂z}{∂t} = 0~\text{. Additionally, the partial derivative of t with respect to itself is 1}~\frac{∂t}{∂t} = 1$.
Thus, the expression simplifies to: $(\frac{∂w}{∂t})_{x, z} = \frac{∂w}{∂y}\frac{∂y}{∂t} + \frac{∂w}{∂t}$.
Besides, from our original equation (i), let’s compute partial derivatives of w: $\frac{∂w}{∂t} = -sin(t), \frac{∂w}{∂y} = 1$.
We use the constraint given in equation t = x - y(ii) and differentiate both sides with respect to t: $\frac{∂x}{∂t}-\frac{∂y}{∂t} = 1$ ⇒[x is constant, $\frac{∂x}{∂t} = 0$] $\frac{∂y}{∂t} = -1$.
Finally, combining all these intermediate calculations, we get: $(\frac{∂w}{∂t})_{x, z} = \frac{∂w}{∂y}\frac{∂y}{∂t} + \frac{∂w}{∂t} = 1·(-1) - sin(t) = -1 -sin(t).$
The goal is to calculate $(\frac{∂w}{∂x})$, which represents the partial derivative of w with respect to t. Then, w is the dependent variable and x is the independent variable. Notice that w, z(dependent), x and y (independent) or w, y (dependent), x and z (independent).
w = x^{2} + y^{2} +z^{2} (i) =[The dependent variables are w and z. We eliminate z using the constraint z = x^{2} + y^{2}(ii)] x^{2} + y^{2} + (x^{2} + y^{2})^{2} = x^{2} + y^{2} + x^{4} + 2x^{2}y^{2} + y^{4}.
$\frac{∂w}{∂x} = 2x +4x^3+4xy^2.$ This is the formula when x and y are the independent variables.
Similarly, where the independent variables are x and z: w = x^{2} + y^{2} +z^{2} (i) =[The dependent variables are z and y. We eliminate y using the constraint z = x^{2} + y^{2}(ii), y^{2} = z -x^{2}] x^{2} + z -x^{2} +z^{2} = z + z^{2}.
$\frac{∂w}{∂x} = 0.$ This is the formula when x and z are the independent variables, and is different from the previous one, what is going on?
w = x^{2} + y^{2} +z^{2} measures the square of the distance from the point (x, y, z) to the origin. The constraint z = x^{2} + y^{2} says that the point (x, y, z) lies on the elliptic paraboloid opening upwards with the vertex at the origin.
The level curves of this surface, which are the intersections of the surface with planes parallel to the xy-plane, are circles centered at the origin. Specifically, the level curve at height z = c is given by the equation x^{2} + y^{2} = c which describes a circle of radius $\sqrt{2}$.
If we take x and y to be independent, then we find $\frac{∂w}{∂x}$ by holding y fixed (e.g., y = 0) and letting x vary. Therefore, with y constant, we are moving along the parabola z = x^{2} in the xz-plane. As we move a point P along this parabola, w which is the square of the distance from P to the origin, charges. We calculate in this case $\frac{∂w}{∂x} = 2x +4x^3+4xy^2,$ e.g., P(1, 0, 1), $\frac{∂w}{∂x} = 2x +4x^3+4xy^2 = 2 +4 + 0 = 6.$
If we take x and z to be independent, then we calculate $\frac{∂w}{∂x}$ by holding z fixed and letting x vary. Since the z-coordinate of P is 1, varying x moves P along a circle x^{2}+y^{2} = 1 in the plane z = 1. In general, With z constant, we consider movement along a circle in the xy-plane. As P moves along this circle, its distance from the origin is constant. Therefore, w, being as it is the square of this distance, do not change at all, so $\frac{∂w}{∂x} = 0$ (Figure 1).
Identify the partial derivatives of z with respect to x and y. Given $z = x^3y+3x^2y^3$, calculate $\frac{∂z}{∂x}, \frac{∂z}{∂y}$: $\frac{∂z}{∂x} = z_x = 3x^2y+6xy^3, \frac{∂z}{∂y} = z_y = x^3+9x^2y^2$
Determine the derivatives of x and y with respect to t. Given $x = 4+t^2, y = 5t^3$, calculate $\frac{∂x}{∂t}, \frac{∂y}{∂t}:$ $\frac{∂x}{∂t} = 2t, \frac{∂y}{∂t} = 15t^2$
Apply the chain rule to calculate $\frac{dz}{dt}$: Using the chain rule: $\frac{dz}{dt}= \frac{∂z}{∂x}\frac{∂x}{∂t} +\frac{∂z}{∂y}\frac{∂x}{∂t} =[\text{Substitute the partial derivatives}] (3x^2y+6xy^3)2t + (x^3+9x^2y^2)15t^2 =[x = 4+t^2, y = 5t^3] [3(4+t^2)^2(5t^3)+6(4+t^2)(5t^3)^3]·2t + [(4+t^2)^3+9(4+t^2)(5t^3)^2]·15t^2$
A triangle is a polygon with three corners and three sides, one of the basic shapes in geometry. The area A of a general triangle is given by: Area = $\frac{1}{2}a·b·sin(θ)$, A = A(a, b, θ), where a and b are the lengths of its two sides, and θ is angle between them.
For a right triangle, we have a special relationship between the sides and the angle: a = b·cos(θ) (Figure B). This introduces the constraint.
We may consider a, b, θ as independent variables: $\frac{∂A}{∂θ} = (\frac{∂A}{∂θ})_{a b}$.
$\text{However, this is not possible because then the triangle would not be a right triangle}, \frac{∂A}{∂θ} = (\frac{∂A}{∂θ})_{a, b} = \frac{1}{2}a·b·cos(θ).$
We want to find the rate of change of the area A with respect to θ, holding one side (say a) constant.
Since a = bcos(θ), we can solve for b: b = b(a, θ) = $\frac{a}{cos(θ)} = asec(θ)$ and by doing so, we keep the right triangle constrain, $(\frac{∂A}{∂θ})_a$
Similarly, we can keep “b” constant, so “a” will change a = a(b, θ) = $\frac{b}{sin(θ)}$ and we keep the right triangle constrain, $(\frac{∂A}{∂θ})_b$, too.
Let’s calculate $(\frac{∂A}{∂θ})_a$ where A = $\frac{1}{2}a·b·sin(θ)$ and a = b·cos(θ).
In this particular case (this is not a general method by any means and we can not be able to use it in many problems), the area of the right triangle is $A = \frac{1}{2}a·b·sin(θ)$
Solve for b in the constraint (b = asec(θ)) and substitute in our main (Area) equation: $A = \frac{1}{2}a^2·sin(θ)·sec(θ) = \frac{1}{2}a^2·\frac{sin(θ)}{cos(θ)} =\frac{1}{2}a^2·tan(θ)$. Finally, we calculate the partial derivative: $(\frac{∂A}{∂θ})_a =[\text{The derivate of tan(θ) is}~ sec^2(θ)] \frac{1}{2}a^2·sec^2(θ).$
Thus, the rate of change of the area with respect to θ, holding “a” constant, is: $(\frac{∂A}{∂θ})_a = \frac{1}{2}a^2·sec^2(θ)$ 🚀
Let’s verify this result using the method of differentials. The general method is to take differentials considering that a is fixed.
From the constraint a = b·cos(θ), we take differentials: da = cos(θ)db -bsin(θ)dθ.
Since a is constant or fixed, da = 0: 0 = cos(θ)db -bsin(θ)dθ ⇒ cos(θ)db =bsin(θ)dθ ⇒ db = btan(θ)dθ
The area is: A = $\frac{1}{2}a·b·sin(θ)$. The total differential dA of a multivariable function A(a, b, θ) is given by dA = $\frac{∂A}{∂a}da + \frac{∂A}{∂b}db + \frac{∂A}{∂θ}dθ$.
Compute the partial derivatives: $\frac{∂A}{∂a} = \frac{1}{2}b·sin(θ), \frac{∂A}{∂b} = \frac{1}{2}a·sin(θ), \frac{∂A}{∂θ} = \frac{1}{2}ab·cos(θ)$.
Substitute these into the differential equation: $dA = \frac{1}{2}·b·sin(θ)·da+\frac{1}{2}·a·sin(θ)·db+\frac{1}{2}abcos(θ)dθ$.
Since da = 0 and db = btan(θ)dθ: $dA = \frac{1}{2}·a·sin(θ)·btan(θ)dθ+\frac{1}{2}abcos(θ)dθ = \frac{1}{2}ab(sin(θ)tan(θ)+cos(θ))dθ = \frac{1}{2}ab(\frac{sin^2(θ)}{cos(θ)}+cos(θ))dθ = \frac{1}{2}ab(\frac{sin^2(θ)+cos^2(θ)}{cos(θ)})dθ = \frac{1}{2}ab(\frac{1}{cos(θ)})dθ = \frac{1}{2}absec(θ)dθ$
$ dA = \frac{1}{2}absec(θ)dθ ⇒ (\frac{∂A}{∂θ})_a$ =[dA = $\frac{∂A}{∂a}da + \frac{∂A}{∂b}db + \frac{∂A}{∂θ}dθ$] $\frac{1}{2}absec(θ)$. Given b = $\frac{a}{cos(θ)}, (\frac{∂A}{∂θ})_a = \frac{1}{2}a^2sec^2(θ)$ 🚀
⋄ Conclusion: We have confirmed using both direct substitution and differentials that the rate of change of the area of a right triangle with respect to the angle θ, holding a constant, is: $(\frac{∂A}{∂θ})_a = \frac{1}{2}a^2sec^2(θ)$
The idea is to write dA in terms of da, db, and dθ, set da = 0 (a = constant), differentiate the constraint, so we can solve for db in terms of dθ and plug these intermediary results into dA so we can get the answer.
Another method to find the rate of change of the area of a right triangle with respect to the angle θ while keeping one side (a) constant is to apply the chain rule.
The chain rule helps us find how a function changes with respect to one variable by considering its dependence on multiple variables. For our problem, we want to find $(\frac{∂A}{∂θ})_a$, the rate of change of A with respect to θ while holding a constant. The chain rule gives us:
$(\frac{∂A}{∂θ})_a = \frac{∂A}{∂θ}(\frac{∂θ}{∂θ})_a + \frac{∂A}{∂a}(\frac{∂a}{∂θ})_a + \frac{∂A}{∂b}(\frac{∂b}{∂θ})_a = A_θ(\frac{∂θ}{∂θ})_a + A_a(\frac{∂a}{∂θ})_a + A_b(\frac{∂b}{∂θ})_a$
Simplify the expression where A = A(a, b, θ), $A_θ = \frac{∂A}{∂θ}$, the partial derivative of A with respect to θ or the rate of change of A with respect to θ when we hold all other variables, namely “a” and “b”, constant. $(\frac{∂θ}{∂θ})_a=1$ because it is the rate of change of θ with respect to itself and $(\frac{∂a}{∂θ})_a = 0$ because a is constant.
The previous equation simplifies to: $(\frac{∂A}{∂θ})_a = \frac{∂A}{∂θ} + \frac{∂A}{∂b}(\frac{∂b}{∂θ})_a$
We need to calculate $(\frac{∂b}{∂θ})_a$ using the constrain a = bcos(θ).
Differentiating both sides with respect to θ and taking into account that a is fixed, $0 = cos(θ)db-bsin(θ)dθ⇒ cos(θ)db = bsin(θ)dθ ⇒ \frac{db}{dθ} = btan(θ).$ Thus, $(\frac{∂b}{∂θ})_a = btan(θ)$
Compute the Partial Derivatives of A, $A = \frac{1}{2}a·b·sin(θ), \frac{∂A}{∂θ} = \frac{1}{2}a·b·cos(θ), \frac{∂A}{∂b} = \frac{1}{2}a·sin(θ)$
Substitute the partial derivatives and $(\frac{∂b}{∂θ})_a$ into the simplified chain rule expression:
$(\frac{∂A}{∂θ})_a = \frac{∂A}{∂θ} + \frac{∂A}{∂b}(\frac{∂b}{∂θ})_a = \frac{1}{2}a·b·cos(θ) + \frac{1}{2}a·sin(θ)(btan(θ)) = \frac{1}{2}a·b·cos(θ) + \frac{1}{2}a·b\frac{sin^2(θ)}{cos(θ)} = \frac{1}{2}a·b(cos(θ)+\frac{sin^2(θ)}{cos(θ)}) = \frac{1}{2}a·b(\frac{cos^2(θ)+sin^2(θ)}{cos(θ)}) = \frac{1}{2}a·b\frac{1}{cos(θ)} = \frac{1}{2}a·b·sec(θ)$.
Since b = asec(θ)⇒$(\frac{∂A}{∂θ})_a = \frac{1}{2}a^2·sec^2(θ)$ 🚀 Using the chain rule, we have confirmed that the rate of change of the area of a right triangle with respect to the angle θ, holding a constant, is: $(\frac{∂A}{∂θ})_a = \frac{1}{2}a^2·sec^2(θ)$
This method, although more involved, systematically breaks down the problem and reinforces the concept of applying the chain rule in multivariable calculus.
Given a function f = f(x, y, z) with the constrain g(x, y, z) = 0, and where y is held constant while x varies, and x = x(y, z).
We aim to calculate $(\frac{∂f}{∂z})_y$, which is the rate of change of f with respect to z while holding y constant.
To calculate $(\frac{∂f}{∂z})_y$ using differentials, we start with the total differential of f: df = $\frac{∂f}{∂x}dx + \frac{∂f}{∂y}dy + \frac{∂f}{∂z}dz = f_xdx + f_ydy + f_zdz$
Since y is constant, dy = 0, so the previous equation simplifies to: $df = f_xdx + f_zdz$.
Next, we consider the constraint g(x, y, z) = 0. We need dx in terms of dz. The total differential of g is: dg = $g_xdx + g_ydy + g_zdz = 0$. Again, since y is constant (dy = 0), and the equation reduces to: $g_xdx + g_zdz = 0$.
Solving for dx, we get $dx = \frac{-g_z}{g_x}dz$ and this equation represents the rate of change of x with respect to z when y is constant, i.e., $(\frac{∂x}{∂z})_y = \frac{-g_z}{g_x}dz$.
Next, we plug or substitute this expression into the differential for df, df = $f_xdx + f_zdz = f_x\frac{-g_z}{g_x}dz + f_zdz = (f_x\frac{-g_z}{g_x} + f_z)dz$.
Therefore, $(\frac{∂f}{∂z})_y = -f_x\frac{g_z}{g_x} + f_z$.
We can also apply the chain rule to find $(\frac{∂f}{∂z})_y$. The chain rule in this context is:
$(\frac{∂f}{∂z})_y = \frac{∂f}{∂x}(\frac{∂x}{∂z})_y+\frac{∂f}{∂y}(\frac{∂y}{∂z})_y+\frac{∂f}{∂z}(\frac{∂z}{∂z})_y$
Give that y is held constant: $(\frac{∂y}{∂z})_y = 0, (\frac{∂z}{∂z})_y=1$. Thus, the chain rule simplifies to: $(\frac{∂f}{∂z})_y = \frac{∂f}{∂x}(\frac{∂x}{∂z})_y+\frac{∂f}{∂z}$
Using the constraint g(x, y, z) = 0, we differentiate g with respect to z⇒ $0 = (\frac{∂g}{∂z})_y = \frac{∂g}{∂x}(\frac{∂x}{∂z})_y+\frac{∂g}{∂y}(\frac{∂y}{∂z})_y+\frac{∂g}{∂z}(\frac{∂z}{∂z})_y =[\text{Similarly,}(\frac{∂y}{∂z})_y = 0, (\frac{∂z}{∂z})_y = 1] \frac{∂g}{∂x}(\frac{∂x}{∂z})_y+\frac{∂g}{∂z} = 0 ⇒0 = \frac{∂g}{∂x}(\frac{∂x}{∂z})_y+\frac{∂g}{∂z} = g_x(\frac{∂x}{∂z})_y+g_z ⇒ (\frac{∂x}{∂z})_y = \frac{-g_z}{g_x}$, and yet again we substitute $(\frac{∂x}{∂z})_y$ back into the chain rule equation, $(\frac{∂f}{∂z})_y = \frac{∂f}{∂x}(\frac{∂x}{∂z})_y+\frac{∂f}{∂z} = \frac{∂f}{∂x}\frac{-g_z}{g_x}+\frac{∂f}{∂z} = f_x\frac{-g_z}{g_x} + f_z$, $(\frac{∂f}{∂z})_y = f_x\frac{-g_z}{g_x} + f_z$.
Both method yield the same result, confirming their consistency. The rate of change of f with respect to z while holding y constant is: $(\frac{∂f}{∂z})_y = f_x\frac{-g_z}{g_x} + f_z$.
By employing these techniques, we reinforce the understanding of handling partial derivatives and constraints in multivariable calculus.