How a Single Derivative Gives Both the Tangent and the Normal to a Curve

The derivative is the slope of a curve at a point, so it gives the tangent line immediately and the perpendicular normal line with one more step. Because it is also a rate, it links how fast one quantity changes to how fast another does.

This lesson covers tangents and normals, angles between curves, related rates, and rates of change of geometric and physical quantities.

**At the point $(x_1,y_1)$ on a curve where the slope is $m=\frac{dy}{dx}$, the tangent is $y-y_1=m(x-x_1)$ and the normal is $y-y_1=-\frac{1}{m}(x-x_1)$, with horizontal and vertical lines as the special cases when m is 0 or undefined.

Worked example.** Find the tangent and normal to $y=x^3-2x+1$ at $(2,5)$.

- $\frac{dy}{dx}=3x^2-2$, which is 10 at $x=2$
- Tangent: $y-5=10(x-2)$, that is $y=10x-15$
- Normal: $y-5=-\frac{1}{10}(x-2)$, that is $x+10y-52=0$

Special cases:

- If $\frac{dy}{dx}=0$, the tangent is $y=y_1$ and the normal is $x=x_1$
- If $\frac{dy}{dx}$ is undefined, the tangent is vertical

An everyday example. A car skidding off a curved road on a wet night moves along the tangent to the road at the point where the tyres lose grip.

The substance. The normal at any point of a circle passes through its centre — a quick check on any circle problem.

**The angle between two curves at a common point is the angle between their tangents there, given by $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$, and the curves cut at right angles when $m_1{m}_2=-1$.

Steps:

- Solve the two equations together to find the point of intersection
- Find each curve's slope at that point
- Apply the angle formula

Worked example.** Find the angle between $y=x^2$ and $y^2=x$ at $(1,1)$.

- For $y=x^2$: $m_1=2x=2$
- For $y^2=x$: $2y\frac{dy}{dx}=1$, so $m_2=\frac{1}{2}$
- $\tan \theta =\left|\frac{2-0.5}{1+1}\right|=0.75$, so $\theta \approx 36.9°$

Worked example 2. Show that $x^2=4y$ and $4y+x^2=8$ cut at right angles at $(2,1)$. Their slopes there are $\frac{x}{2}=1$ and $-\frac{x}{2}=-1$, whose product is $-1$.

An everyday example. Two curving roads meeting at a junction cross at the angle between their directions at the meeting point.

The substance. The same pair of curves can meet at different angles at different points — $y=x^2$ and $y^2=x$ also meet at the origin, where they cross at right angles.

In a related-rates problem, write an equation linking the changing quantities, differentiate both sides with respect to time using the chain rule, and substitute the known values only after differentiating.

Steps:

- Name the variables and note which rates are given and which is required
- Write the equation connecting the variables
- Differentiate with respect to t
- Substitute the values at the instant asked about

Worked example. A stone dropped into a still pond makes circular ripples whose radius grows at 5 cm/s. How fast is the disturbed area growing when the radius is 10 cm?

$$A=\pi r^2\Rightarrow \frac{dA}{dt}=2\pi r\frac{dr}{dt}=2\pi \times 10\times 5=100\pi \approx 314{\text{ cm}}^2\text{/s}$$

Worked example 2. A 5 m ladder slides down a wall, its foot moving away at 0.3 m/s. When the foot is 3 m from the wall the top is 4 m high, and from $x^2+y^2=25$,

$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\Rightarrow \frac{dy}{dt}=-\frac{3\times 0.3}{4}=-0.225\text{ m/s}$$

so the top slides down at 0.225 m/s.

An everyday example. Water poured into a village storage tank raises the level at a rate that depends on both the pouring rate and the tank's shape.

The substance. Substituting before differentiating is the classic error — putting $r=10$ into $A=\pi r^2$ first turns it into a constant, whose derivative is wrongly zero.

**Any quantity that depends on another — volume on radius, cost on output, or position on time — has a rate of change given by its derivative, so velocity is $\frac{ds}{dt}$, acceleration is $\frac{dv}{dt}$, and marginal cost is $\frac{dC}{dx}$.

Worked example (geometry).** A spherical balloon's volume increases at 100 cm³/s. How fast is its radius increasing when the radius is 5 cm?

$$V=\frac{4}{3}\pi r^3\Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=\frac{100}{4\pi \times 25}=\frac{1}{\pi }\approx 0.318\text{ cm/s}$$

Worked example (physics). A particle has position $s=t^3-6t^2+9t$ metres. Its velocity is $v=3t^2-12t+9$ and its acceleration is $a=6t-12$. It is momentarily at rest when $v=0$, at $t=1$ s and $t=3$ s, and at $t=4$ s its acceleration is 12 m/s².

Worked example (economics). If making x units costs $C(x)=0.005x^3-0.02x^2+30x+500$ rupees, the marginal cost at $x=10$ is $C^{\prime }(10)=0.015(100)-0.04(10)+30=31.1$ rupees per unit.

An everyday example. A two-wheeler's speedometer shows the rate of change of distance, while hard braking produces a large negative rate of change of speed.

The substance. A negative rate means the quantity is decreasing — the sign carries information, so it must never be dropped.