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“I intend to live forever. So far, so good.” – Steven Wright
A function of two variables $f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ assigns a unique real number to each ordered pair in its domain, e.g., $Area = \frac{1}{2}bh$, $z = 2x + 3y$, and $f(x, y) = x^2 + y^2$.
Partial derivatives are derivatives of a function of multiple variables, say $f(x_1, x_2, \cdots, x_n)$, 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 $z = f(x,y)$ 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 $z = f(x,y)$ and the plane $x = x_0$.
The directional derivative $\frac{dw}{ds}\bigg\vert{}_{\vec{u}}$ is the rate of change of the function $w$ as we move in the direction of a unit vector $\vec{u}$:
$\frac{dw}{ds}\bigg\vert{}_{\vec{u}} = \nabla w \cdot \vec{u} = \vert{}\nabla w\vert{} \cdot \vert{}\vec{u}\vert{}\cos(\theta) = \vert{}\nabla w\vert{}\cos(\theta)$ where $\theta$ is the angle between the gradient and the given unit vector $\vec{u}$.
The gradient vector is defined as $\nabla w = \langle w_x, w_y, w_z \rangle = \langle \frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z} \rangle$. Because there is zero change in the function’s value along a level curve or level surface, the rate of change is zero, meaning the gradient is always perpendicular (orthogonal) to the level curves.
Let $w = f(x, y, z)$ where $x = x(t)$, $y = y(t)$, and $z = z(t)$ are parametrised functions of another variable $t$. The derivative of $w$ with respect to $t$ is calculated via the Chain Rule: $\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}$
This can be neatly rewritten using the dot product of the gradient and the velocity vector: $\frac{dw}{dt} = \nabla w \cdot \frac{d\vec{r}}{dt}$ where $\frac{d\vec{r}}{dt} = \langle \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \rangle$.
Identify the partial derivatives of z with respect to x and y. Given $z = x^3y+3x^2y^3$, calculate $\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}$: $\frac{\partial z}{\partial x} = z_x = 3x^2y+6xy^3, \frac{\partial z}{\partial y} = z_y = x^3+9x^2y^2$
Determine the derivatives of x and y with respect to t.
Because $x$ and $y$ depend solely on a single variable $t$, we use ordinary derivatives $\frac{d}{dt}$ rather than partial derivatives $\frac{\partial}{\partial t}$.
Given $x = 4+t^2, y = 5t^3$: $\frac{dx}{dt} = 2t$, $\frac{dy}{dt} = 15t^2$.
Apply the chain rule to calculate $\frac{dz}{dt}$: $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$
Substitute the computed derivatives into the formula: $\frac{dz}{dt} = (3x^2y + 6xy^3)(2t) + (x^3 + 9x^2y^2)(15t^2)$
Substitute the original constraint equations ($x = 4+t^2$, $y = 5t^3$) to express the final answer entirely in terms of $t$: $\frac{dz}{dt} = [3(4+t^2)^2(5t^3) + 6(4+t^2)(5t^3)^3]2t + [(4+t^2)^3 + 9(4+t^2)^2(5t^3)^2]15t^2$
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: $A(a, b, \theta) = \frac{1}{2}ab\sin(\theta)$, where $a$ and $b$ are side lengths and $\theta$ is the angle between them.
For a right triangle where the adjacent side $a$ and hypotenuse $b$ enclose angle $\theta$, we have a special relationship between the sides and the angle: $a = b\cos(\theta)$ (Figure B). This introduces the constraint.
We want to find the rate of change of the area $A$ with respect to $\theta$, holding side $a$ constant: $\left(\frac{\partial A}{\partial \theta}\right)_a$. Because treating $a, b, \text{and } \theta$ as purely independent variables would break the right triangle constraint, we must use substitution or differentials.
Method 1: Direct Substitution
Since $a = b\cos(\theta)$, we can solve the constraint for $b$: $b = b(a, \theta) = \frac{a}{\cos(\theta)} = a\sec(\theta)$ and by doing so, we keep the right triangle constraint which validates computing $(\frac{\partial A}{\partial \theta})_a$ by substitution.
Similarly, we can keep “b” constant, so “a” will change $a = a(b, \theta) = \frac{b}{\sin(\theta)}$ and we keep the right triangle constraint, $(\frac{\partial A}{\partial \theta})_b$, too.
Substitute $b$ into our main (area) equation: $A = \frac{1}{2}a(a\sec(\theta))\sin(\theta) = \frac{1}{2}a^2\frac{\sin(\theta)}{\cos(\theta)} = \frac{1}{2}a^2\tan(\theta)$
Differentiate with respect to $\theta$ (treating $a$ as a constant): $\left(\frac{\partial A}{\partial \theta}\right)_a = \frac{1}{2}a^2\sec^2(\theta)$
Thus, the rate of change of the area with respect to $\theta$, holding “a” constant, is: $\left(\frac{\partial A}{\partial \theta}\right)_a = \frac{1}{2}a^2\sec^2(\theta)$
Method 2: General Method Using Differentials
This confirms the result by expressing $dA$ in terms of $da, db, \text{and } d\theta$, manually locking $a$ by setting $da = 0$ (we are taking differentials considering that a is fixed).
Differentiate the constraint $a = b\cos(\theta)$: $da = \cos(\theta)db - b\sin(\theta)d\theta$
Since $a$ is held constant, $da = 0$:
$0 = \cos(\theta)db - b\sin(\theta)d\theta \implies db = b\tan(\theta)d\theta$
Compute the total differential of the original area equation $A = \frac{1}{2}ab\sin(\theta)$.
The total differential $dA$ of a multivariable function $A(a, b, \theta)$ is given by $dA = \frac{\partial A}{\partial a}da + \frac{\partial A}{\partial b}db + \frac{\partial A}{\partial \theta}d\theta$
Compute the partial derivatives: $dA = \left[\frac{1}{2}b\sin(\theta)\right]da + \left[\frac{1}{2}a\sin(\theta)\right]db + \left[\frac{1}{2}ab\cos(\theta)\right]d\theta$
Substitute $da = 0$ and $db = b\tan(\theta)d\theta$ into the total differential: $dA = 0 + \frac{1}{2}a\sin(\theta)(b\tan(\theta)d\theta) + \frac{1}{2}ab\cos(\theta)d\theta$
$dA = \frac{1}{2}ab \left( \sin(\theta)\tan(\theta) + \cos(\theta) \right) d\theta$
$dA = \frac{1}{2}ab \left( \frac{\sin^2(\theta)}{\cos(\theta)} + \frac{\cos^2(\theta)}{\cos(\theta)} \right) d\theta$
$dA = \frac{1}{2}ab \left( \frac{1}{\cos(\theta)} \right) d\theta = \frac{1}{2}ab\sec(\theta)d\theta$
Extract the constrained partial derivative:
Once we’ve expressed $dA$ in terms of $d\theta$ alone (with the constraint applied) $dA = \left(\frac{\partial A}{\partial \theta}\right)_a d\theta$. In other words, under the constraint $a = \text{constant}$, any small change in $A$ due to a small change in $\theta$ is described by: $dA = \left(\frac{\partial A}{\partial \theta}\right)_a d\theta$, the factor multiplying $d\theta$ is exactly $\left(\frac{\partial A}{\partial \theta}\right)_a$: $\left(\frac{\partial A}{\partial \theta}\right)_a = \frac{1}{2}ab\sec(\theta)$
Finally, substitute $b = a\sec(\theta)$ to reach the final answer: $\left(\frac{\partial A}{\partial \theta}\right)_a = \frac{1}{2}a^2\sec^2(\theta)$
Conclusion. Both direct substitution and the method of differentials confirm that the rate of change of the area of a right triangle with respect to the angle $\theta$, holding $a$ constant, is $\frac{1}{2}a^2\sec^2(\theta)$.
Method 3: The Multivariable Chain Rule
Another method to find the rate of change of the area of a right triangle with respect to the angle $\theta$ while keeping one side ($a$) constant is to apply the chain rule.
The chain rule tracks how a function changes with respect to one variable by considering its dependence on multiple variables. For our problem, we want to find $(\frac{\partial A}{\partial \theta})_a$, the rate of change of $A(a, b, \theta)$ with respect to θ while holding a constant.
The chain rule gives us: $(\frac{\partial A}{\partial \theta})_a = \frac{\partial A}{\partial \theta}(\frac{\partial \theta}{\partial \theta})_a + \frac{\partial A}{∂a}(\frac{∂a}{\partial \theta})_a + \frac{\partial A}{\partial b}(\frac{\partial b}{\partial \theta})_a = A_θ(\frac{\partial \theta}{\partial \theta})_a + A_a(\frac{∂a}{\partial \theta})_a + A_b(\frac{\partial b}{\partial \theta})_a$
We can immediately simplify this equation: $\left(\frac{\partial \theta}{\partial \theta}\right)_a = 1$ (the rate of change of $\theta$ with respect to itself). Besides, $\left(\frac{\partial a}{\partial \theta}\right)_a = 0$ (because $a$ is held constant).
The previous equation reduces to: $(\frac{\partial A}{\partial \theta})_a = \frac{\partial A}{\partial \theta} + \frac{\partial A}{\partial b}(\frac{\partial b}{\partial \theta})_a$
Calculate the constrained derivative of $b$:
We need to calculate $(\frac{\partial b}{\partial \theta})_a$ using the constraint $a = b\cos(\theta)$.
Differentiating both sides with respect to $\theta$ and taking into account that a is fixed, $0 = \cos(\theta)db - b\sin(\theta)d\theta \implies \cos(\theta)db = b\sin(\theta)d\theta \implies \frac{db}{d\theta} = b\tan(\theta)$
Therefore, $\left(\frac{\partial b}{\partial \theta}\right)_a = b\tan(\theta)$.
Compute the partial derivatives of $A$:
Given $A = \frac{1}{2}ab\sin(\theta)$: $\frac{\partial A}{\partial \theta} = \frac{1}{2}ab\cos(\theta)$ and $\frac{\partial A}{\partial b} = \frac{1}{2}a\sin(\theta)$
Substitute all parts back into the simplified chain rule: $\left(\frac{\partial A}{\partial \theta}\right)_a = \frac{1}{2}ab\cos(\theta) + \left[ \frac{1}{2}a\sin(\theta) \right]\left[ b\tan(\theta) \right]$
$\left(\frac{\partial A}{\partial \theta}\right)_a = \frac{1}{2}ab\cos(\theta) + \frac{1}{2}ab\frac{\sin^2(\theta)}{\cos(\theta)}$
$\left(\frac{\partial A}{\partial \theta}\right)_a = \frac{1}{2}ab \left( \frac{\cos^2(\theta) + \sin^2(\theta)}{\cos(\theta)} \right) = \frac{1}{2}ab\left(\frac{1}{\cos(\theta)}\right) = \frac{1}{2}ab\sec(\theta)$
Since the constraint tells us $b = a\sec(\theta)$, substituting this gives the final confirmed result: $\left(\frac{\partial A}{\partial \theta}\right)_a = \frac{1}{2}a^2\sec^2(\theta)$
Conclusion. 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 $f = f(x, y, z)$ subject to a constraint $g(x, y, z) = 0$, we aim to calculate the rate of change of $f$ with respect to $z$ while holding $y$ constant, where $x$ is free to vary as a function of $y$ and $z$.
Method 1: Using Differentials
Write the total differential of $f$: $df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz = f_xdx + f_ydy + f_zdz$
Since $y$ is constant, $dy = 0$. The equation simplifies to: $df = f_xdx + f_zdz$
Use the constraint $g(x,y,z) = 0$ to find $dx$:
The total differential of $g$ is: $dg = g_xdx + g_ydy + g_zdz = 0$
Again, set $dy = 0$: $g_xdx + g_zdz = 0 \implies[\text{Solving for dx}] dx = -\frac{g_z}{g_x}dz$
This equation represents the rate of change of x with respect to z when y is constant, i.e. $\left(\frac{\partial x}{\partial z}\right)_y$.
Plug or substitute $dx$ into the differential for $df$:
$df = f_x\left(-\frac{g_z}{g_x}dz\right) + f_zdz = \left(f_z - f_x\frac{g_z}{g_x}\right)dz$
Therefore: $\boxed{\left(\frac{\partial f}{\partial z}\right)_y = f_z - f_x\frac{g_z}{g_x}}$. This assumes $g_x \ne 0$.
Method 2: Using the Chain Rule
Apply the chain rule directly to $f$: $\left(\frac{\partial f}{\partial z}\right)_y = \frac{\partial f}{\partial x}\left(\frac{\partial x}{\partial z}\right)_y + \frac{\partial f}{\partial y}\left(\frac{\partial y}{\partial z}\right)_y + \frac{\partial f}{\partial z}\left(\frac{\partial z}{\partial z}\right)_y$
Simplify the trivial derivatives:
Because $y$ is held constant, $\left(\frac{\partial y}{\partial z}\right)_y = 0$. Naturally, $\left(\frac{\partial z}{\partial z}\right)_y = 1$.
The formula becomes:$\left(\frac{\partial f}{\partial z}\right)_y = f_x\left(\frac{\partial x}{\partial z}\right)_y + f_z$
Differentiate the constraint to find the missing term:
Apply the same chain rule logic to the constraint equation $g(x, y, z) = 0$: $0 = \left(\frac{\partial g}{\partial z}\right)_y = g_x\left(\frac{\partial x}{\partial z}\right)_y + g_y(0) + g_z(1)$
$0 = g_x\left(\frac{\partial x}{\partial z}\right)_y + g_z \implies \left(\frac{\partial x}{\partial z}\right)_y = -\frac{g_z}{g_x}$
Substitute back into the simplified chain rule:
$\left(\frac{\partial f}{\partial z}\right)_y = f_x\left(-\frac{g_z}{g_x}\right) + f_z$
Conclusion. Both methods yield the same result, confirming their consistency.
By employing these techniques, we reinforce the understanding of handling partial derivatives and constraints in multivariable calculus.
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