MATH 145 – Calculus I
October 16, 2025
So far, we’ve worked solely with functions that can be described explicitly in terms of \(x\), i.e. \(f(x) = x^3 + 2x^2 + 3x + 4\).
But not all curves can be described this way.
The most obvious example is a circle. Here is the plot of \(x^2 + y^2 = 16\), the circle centered at the origin with radius \(4\):
Here is the plot of \(x^3 - y^3 = 6xy\):
What if we want to find the rate of change of the function \(x^2 + y^2 = 16\) at the point \(\left( 2\sqrt{2}, 2\sqrt{2} \right)\)?
Consider the circle \(x^2 + y^2 = 16\).
We can’t differentiate the normal way, but what if we think of \(y\) as a implicit function of \(x\), but which we can differentiate?
Now we can’t find the derivative all the way, we won’t know the derivative, \(\dfrac{dy}{dx}\), directly (yet). But we can use the Chain Rule!
\[\begin{align*} \dfrac{d}{dx} \Big[ x^2 + y^2 \Big] &= \dfrac{d}{dx} [16] \\ \dfrac{d}{dx} \Big[ x^2 \Big] + \dfrac{d}{dx} \Big[ y^2 \Big] &= 0 \\ 2x + 2y \dfrac{dy}{dx} &= 0 \end{align*}\]
As by the Chain Rule: \[ \dfrac{d}{dx} \Big[ y(x)^2 \Big] = 2 (y(x))^1 \frac{dx}{dy} \]
From here, solve for \(\dfrac{dy}{dx}\) to find the form of the derivative: \[\begin{align*} 2x + 2y \dfrac{dy}{dx} &= 0 \\ 2y \dfrac{dy}{dx} &= -2x \\ \dfrac{dy}{dx} &= -\dfrac{2x}{2y} = -\dfrac{x}{y} \end{align*}\]
What is the slope of the tangent line at \(\left( 2, 2\sqrt{3} \right)\)?
Examples:
Consider the curve given by the equation \(2y^3 + y^2 - y^5 = x^4 - 2x^3 + x^2\).
Consider a right triangle with the three sides changing. How can we relate the rates of change of the lengths of the three sides?
The three sides are related through the Pythagorean Theorem \(a^2 + b^2 = c^2\).
Each of the lengths of the three sides of the triangle are changing as a function of time, but are also tied to each other.
So what if we derive all three in terms of the time, \(t\)?
\[\begin{align*} \dfrac{d}{dt} \Big[ a^2 + b^2 \Big] &= \dfrac{d}{dt} \Big[ c^2 \Big] \\ \dfrac{d}{dt} \Big[ a^2 \Big] + \dfrac{d}{dt} \Big[ b^2 \Big] &= \dfrac{d}{dt} \Big[ c^2 \Big] \\ 2a \dfrac{da}{dt} + 2b \dfrac{db}{dt} &= 2c \dfrac{dc}{dt} \end{align*}\]
Now if we know the rates of change of two of the sides, we can tell the rate of change of the third definitively.
Water is draining from the bottom of a cone-shaped funnel at a rate of \(0.03 \frac{ft^3}{sec}\). The height of the funnel is \(2\) feet and the radius at the top of the funnel is \(1\) foot. At what rate is the height of the water in the funnel changing when the height of the water is \(\dfrac{1}{2}\) feet?