Differentials

Tangent Line Approximations

Tangent line approximations are used in many ways in calculus. In Newton's Method, the tangent line is taken to find the approximation. Recall that the formula of the tangent line at (c, f(c)) is 

$$y - f(c) = f'(c)(x-c)\text{, and this is written in point slope form}$$ $$\text{Equivalent: } y = f(c) + f'(c)(x-c)$$

Differentials

So, let's say you have this tangent line: y - f(c) =  f'(c)(x-c). In that equation, you want to find out (x, f(x)) and you're given (c, f(c)). So, x-c is the change in x, or Δx. The change in y, or Δy is y - f(c). So, you can rewrite the equation as Δy = f'(c)Δx. Usually, this isn't written with Δs. Instead, it's usually written as dy = f'(x)dx, and this is because dx is explicitly nonzero. Also, dy is an approximation, and Δy is the actual value. Note that this formula assumes that a function is locally linear, meaning that as two points get closer and closer together, the function becomes very much like a line.

It should also be noted that dy/dx = f'(x), which are equivalent notations of taking a derivative. This works because the derivative is just the instantaneous slope, which is Δy/Δx. If you know for sure that Δx != 0, then you can multiply by dx. 

Compare Δy and dy

$$\text{If } f(x) = x^2 \text{, find dy when } x = 1 \text{and } dx = 0.01. \text{ Compare with } \Delta{}y \text{ for x = 1 and dx = 0.01}$$ Recall dx is what you estimate, and $\Delta{}y$ is the actual change in y.

dy = f'(x)dx = (2x)(0.01) = 0.02x = 0.02(1) = 0.02

Δy = f(1.01) - f(1) = 1.0201 - 1 = 0.0201.

So, they ended up being very close, and when you're dealing with a very small values of dx, differentials become more and more accurate.

Estimating Errors

If you're measuring a square in cm, and you know the measurement is accurate to within 5 mm, or .5 cm. So, the measurement is x cm +- .5 cm. The error is affected differently though. The propagated error in that is (x +/- .5)^2 - x^2 = Δy = x +/- 0.25. 

Using Differentials

Let's say the radius of a ball is supposed to be 1 inch, and your measurement is correct to within 0.01 inch, estimate the propagated error in the volume of the ball bearing. 

Firstly, we let x = 1, and Δx = +/-0.01.

So, ΔV ≈ dV, dV = 4πx²(dx)

(4πx² is the derivative of the volume of a sphere)

dx is +/- 0.01, and x = 1, so dV = 4π(1)^2 * (+/- 0.01)

dV = 0.1256 in³

That's the propagated error.

In order to find out whether the propagated error is big or small, you divide by the area. In this case, you determine that by divided the propagated error by the actual value.

0.1256/(4/3 * π) = 0.02998, or 2.998%

That's the relative error.

Estimating Values

If you wanted to approximating $\sqrt{16.5}$, you could use differentials. $$\mathrm{d}y = f'(x) \mathrm{d}x, \mathrm{d}y = \frac{1}{2\sqrt{x}}(0.5)$$ $$\mathrm{d}y = \frac{1}{2\sqrt{16}}(0.5) = \frac{1}{8}*\frac{1}{2} = \frac{1}{16}$$ $$\sqrt{16} + dy \approx \sqrt{16.5}$$ $$4.0625 \approx \sqrt{16.5}$$