Apply differentiation to connected rates of change, small increments and approximations

Calculus: Differentiation & Rates of Change

1. What is Differentiation?

Differentiation is the process of finding the instantaneous rate of change of a function. In simpler terms, it tells you how fast something is changing at a specific point. Imagine driving a car: the speedometer shows your speed at that exact moment. The derivative is that speedometer for any mathematical function.

If you have a function \(f(x)\), its derivative is written as \(f'(x)\) or \(\frac{df}{dx}\). It represents the slope of the tangent line to the curve at any point \(x\).

2. Basic Differentiation Rules

• Power Rule: If \(f(x)=x^n\), then \(f'(x)=n\,x^{\,n-1}\). 📐
• Constant Rule: If \(f(x)=c\) (where \(c\) is a constant), then \(f'(x)=0\). ❌
• Sum/Difference Rule: \((f\pm g)' = f' \pm g'\). ➕➖
• Product Rule: \((fg)' = f'g + fg'\). 🔗
• Quotient Rule: \(\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}\). ⚖️
• Chain Rule: If \(y = f(g(x))\), then \(y' = f'(g(x)) \cdot g'(x)\). 🔁

3. Connected Rates of Change

Many real‑world problems involve two or more quantities that are linked. By differentiating the relationship between them, we can find how one quantity changes with respect to another.

Example: Water Tank

A tank is being filled at a rate of \( \frac{dV}{dt} = 5\,\text{m}^3/\text{min}\). The tank’s radius \(r\) is changing as water fills it. The volume of a cylinder is \(V = \pi r^2 h\). If the height \(h\) is constant, differentiate to find \( \frac{dr}{dt}\):

1. Start with \(V = \pi r^2 h\).
2. Differentiate both sides w.r.t. \(t\): \( \frac{dV}{dt} = 2\pi r h \frac{dr}{dt}\).
3. Solve for \( \frac{dr}{dt}\): \( \displaystyle \frac{dr}{dt} = \frac{1}{2\pi r h}\frac{dV}{dt}\).

Plug in the numbers to find the rate at which the radius is increasing. 🎈

Example: Balloon Rising

A helium balloon rises at a rate of \( \frac{dh}{dt} = 0.5\,\text{m/s}\). The volume of the balloon is \(V = \frac{4}{3}\pi r^3\). Find how fast the radius is changing when \(r = 0.5\,\text{m}\):

1. Relate \(h\) and \(V\) via the density of helium: \(V = k h\) (where \(k\) is a constant).
2. Differentiate: \( \frac{dV}{dt} = k \frac{dh}{dt}\).
3. But \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\).
4. Set them equal: \(4\pi r^2 \frac{dr}{dt} = k \frac{dh}{dt}\).
5. Solve for \( \frac{dr}{dt}\). 🎈

4. Small Increments & Approximations

When changes are tiny, we can approximate the change in a function using its derivative. This is called a linear approximation.

If \(f(x)\) is differentiable at \(x=a\), then for a small \(\Delta x\): $$\Delta f \approx f'(a)\,\Delta x.$$ The term \(f'(a)\,\Delta x\) is called the differential \(df\).

Example: Estimate \(\sqrt{10.1}\). Let \(f(x)=\sqrt{x}\), \(a=10\), \(\Delta x=0.1\).
Compute \(f'(x)=\frac{1}{2\sqrt{x}}\). At \(x=10\), \(f'(10)=\frac{1}{2\sqrt{10}}\approx0.1581\).
Then \(\Delta f \approx 0.1581 \times 0.1 = 0.0158\).
So \(\sqrt{10.1}\approx f(10)+\Delta f = 3.1623+0.0158\approx3.1781\). ??

5. Examination Tips & Practice Questions

6. Quick Reference Table