Derivative as a rate measurer

Question

I studied about the application of derivatives as they help in measuring rate of change. For example :-

Let $A$ be area of a circle of radius $r$

$$A = \pi \cdot r^2$$

then

$$\frac {dA}{dr}= 2 \pi \cdot r $$

Suppose we have to find rate of change of area w.r.t to radius at $r = 5 \text{ cm}$ .

Then

$$\left(\frac {dA}{dr}\right)_{r=5}= 10 \pi\text{ cm}^2/\text{cm} $$

My question is:

Does our final answer mean that when the radius of the circle changes from $5 \text{cm}$ to $6 \text{cm}$, the change in area is equal to $ 10\pi\text{cm}^2 $, i.e., the area of circle at $6 \text{cm} = \text{area of circle at }5 \text{cm} + 10\pi\ \text{cm}^2 $?

Answer 1

We should write

$$\frac {dA}{d\color{red}r}= 2 \pi \cdot r$$

No of course our final answer doaes not mean that when radius of circle changes from $5 cm$ to $6 cm$ then the change in area is equal to $10\pi\ cm^2$, indeed derivatives are linear approximation and, since in that case the relationship between $A$ and $r$ is not linear, it means that

$$\Delta A \approx 10\pi \Delta r$$

and the approximation becomes better for smaller $\Delta r$.

Answer 2

No, but you are on the right track.

The derivative you have found tells you that when the radius changes from $5$cm to $(5+h)$cm and $h$ is small then the area changes by approximately $10\pi h$ square cm.

In your example you are taking $h=1$, which is not small, so the approximation isn't even close.

Answer 3

No, but the intuition is correct.

$\frac{dA}{dr}$ is the instantaneous rate of change at radius $r$. It means that if your radius is initially $r_0$ (say, 5cm), and you increase the radius by an infinitesimally small amount $dr$, the change in area will be given by $\frac{dA}{dr}$ evaluated at $r_0$. At the new point $r_0 + dr$, the rate of change of area might be different (as it is in this case, where $\frac{dA}{dr}$ depends on $r$. However, if your change in radius is small, your idea could be a good approximation for the change in area.

However, a larger change in area can be viewed as an aggregation of many small infinitesimal changes to $r$. This can be expressed as an integral, and so, the area change when you change the radius from $r_0$ to $r_0+1$ is given by $$\int_{r_0}^{r_0+1} \left(\frac{dA}{dr}\right) dr$$

It could be the case for some functions that the derivative is independent of $r$. In such a case, your idea will give exactly the correct answer, as the rate of change will not change with $r$. An example of such a function would be the rate of change of circumference of a circle with respect to the radius.

Answer 4

Yes it is true, but only for variations such that the linear approximation is acceptable.

For an increase of the radius of $0.001$ cm, the increase in area is approximated by$10\cdot\pi\cdot0.001=0.0314159265\cdots$

Compare to the exact value $\pi(5.001^2-5^2)=0.0314190681\cdots$

In green, the linear approximation.

Answer 5

Let$f: \mathbb{R} \rightarrowtail \mathbb{R}$ be a function, and $a \in \text{Int}(\text{dom}(f))$. Then the following statements are equivalent:

• $f$ is differentiable at $a$
• $\exists \xi \in \mathbb{R}$ and $\exists \eta: \text{dom}(f) \to \mathbb{R}$ with $\lim\limits_{x \to a} \frac{\eta(x)}{x-a}=0$, so that $$f(x)=f(a)+\xi(x-a)+\eta(x)$$ and $\xi=f'(a)$.

I think this equivalent definition of differentiability is more spectacular (and it's easier to generalize to the $\mathbb{R}^n \rightarrowtail \mathbb{R}^m$ case), because we can intuitively see that $\eta(x)$ is really small around $a$, so for $x \approx a$ we have that $$f(x) \approx f(a) + \xi(x-a)$$ In your case, if we let $A(r):=r^2 \pi$, we have that $A'(r)=2r \pi$ $$A(r)=A(r_0)+\xi(r-r_0)+\eta(r)$$ And of course, $\xi=A'(r_0)=2r_0\pi$: $$A(r)=A(r_0)+2r_0(r-r_0)\pi+\eta(r)$$ $$A(r)=r_0^2 \pi+2r_0(r-r_0)\pi+\eta(r)$$ $$A(r)=r_0^2 \pi+2r_0r\pi-2r_0^2 \pi+\eta(r)$$ $$A(r)=2rr_0 \pi-r_0^2 \pi + \eta(r) \tag{1}$$ If we want to get $\eta(r)$: $$r^2 \pi=2rr_0 \pi-r_0^2 \pi + \eta(r)$$ $$r^2 \pi+r_0^2\pi-2r r_0 \pi=\eta(r)$$ $$\eta(r)=(r-r_0)^2 \pi$$ Using some numbers, we have that $A(5)=25 \pi$, $A(5.1)=26.01 \pi$, $A(50)=2500 \pi$. If we try to approximate $A(5.1)$ and $A(50)$ with $A(5)$ and $A'(5)$, we can use the first equation (without $\eta(r)$): $$A(r) \approx 10 r \pi - 25 \pi$$ we get that $A(5.1) \approx 26 \pi$ and $A(50) \approx 475 \pi$, while $\eta(5.1)=0.01 \pi$ and $\eta(50)=2025 \pi$. As you can see, it's quite good for $5.1$, but it's terrible for $50$.

This kind of linear approximation is really useful in physics, for example when you are dealing with a complicated system, you can't really solve the differential equation(s), but you can make the assumption that the change in the sought function (for example, the displacement-time function of a ball) is really small, so you can linearize the equation, and (probably) solve it.

Answer 6

All the previous answers explained the problem quite well. But according to your comment on these answer, you still have trouble understanding when to your approximation holds. I'll try to explain what values of h are acceptable to use your method.

Let's call $A(r)$ the area of a circle of radius $r$. We have $A_{real}=\color{blue}{\pi*r²}$

If $h$ is small enough, you can have $A_{approx}(r_0+h) \approx \color{red}{A(r_0)+2*\pi*r_0*h}$

But keep in mind that the real value of $A(r_0+h)$ is indeed:

\begin{align}A_{real}(r_0+h)=\color{blue}{\pi*(r_0+h)²}\\ & =\pi*({r_0}²+2r_0h+h²)\\ & =\color{blue}{\pi*{r_0}²}+2*\pi*r_0*h+\pi*h²\\ & =\color{red}{A_{real}(r_0)+2*\pi*r_0*h}+\pi*h² \\ & =\color{red}{A_{approx}(r_0+h)} +\pi*h² \end{align}

From there, the relative error between you result and the real one is:

$$E=\frac{A_{real}(r_0+h)-A_{approx}(r_0+h)}{A_{real}(r_0+h)}$$ $$E=\frac{\pi*h²}{\pi*(r_0+h)²}$$ $$E=\frac{h²}{(r_0+h)²}$$

So, you can see that as $h/r_0 \rightarrow 0$, the error $E \rightarrow 0$ too.

This means that as soon as $h/r_0$ is small enough, so is your error. Let see it this way, if you change 1cm in diameter on a circle of 10cm, and calculate the new area with your initial idea, your error will be more important that if you change 1cm on a 1m diameter circle. What only matter in this case is that the change is small enough versus the initial diameter.

The problem in your question is that you chose $h=1$ and $r_0=5$. You'll need $h$ to be smaller or $r_0$ to be greater

I made some calculation for circles of radius 5cm, 10cm, and 1m ($r_0=5,10,100$), with a changing radius of 1cm, you have errors of respectively: $2.7\%$,$0.8\%$ and $0.01\%$ between the real area and the one calculated from your formula.

Answer 7

To say that $\left.\dfrac{dA}{dr}\right|_{r\,=\,5\text{ cm}} = 10\pi\text{ cm}$ means that if you multiply the rate of change of $r,$ at the point at which $r=5\text{ cm},$ by $10\pi\text{ cm},$ you get the rate of change of $A$ at that point. But that does not mean that if $r$ grows by $1\text{ cm},$ then $A$ will increase by $10\pi\text{ cm}^2,$ because it won't continue changing at that same rate as $r$ changes.

Generally if $\left.\dfrac{dy}{dx}\right|_{x\,=\,a} = b$ means that at the point where $x=a,$ the dependent variable $y$ is changing $a$ times as fast as the independent variable $x.$