# Related rates, area of a square

I have no idea how to do these problems anymore.

Each side of a square is increasing at a rate of 6 units per second at what rate is the area of the square increasing when the area o the square is 16units squared.

I have $s' = 6$

and $area = 16$

$area = s^2$

I am not sure how to proceed from here.

-

Let $A(t)$ be the area of the square at time $t$, let $\ell(t)$ be the length of the side of the square at time $t$.

We are given the following information: the rate at which the length of the side (that is, $\ell(t)$) changes is 6 units per second. This is a way of giving you information about the derivative of $\ell$. That is, it's telling you that the rate of change of $\ell$ with respect to time (why time? because it's "per second") is $6$ units per second. In symbols, it's telling you that" $$\frac{d\ell}{dt} = 6\text{ units/second}$$ We are asked to find how fast the area is changing when the area is 16 square units. "How fast" is a dead giveaway that they are asking you for the rate of change of the area; that is, they are asking you for the value of $\frac{dA}{dt}$, the derivative of the area, when the area is equal to 16. In symbols, they are asking you to find: $$\frac{dA}{dt}\Bigm|_{A=16}.$$

The first step, as in any related rates problem, is to find an equation that relates the quantities involved, in this case the length of the side $\ell$ and the area $A$. The relation between $A$ and $\ell$ is given by: $$A = \ell^2.$$ (Geometry: the area of a square is the square of the length of the side)

Once you have the relation between the original quantities, you take derivatives (implicitly) with respect to $t$ to obtain a relation between the quantities and their rates of change: \begin{align*} A &= \ell^2\\ \frac{d}{dt} (A) &= \frac{d}{dt}(\ell^2)\\ \frac{dA}{dt} &= 2\ell\frac{d\ell}{dt}. \end{align*}

Once you have a relation between the quantities and the rates of change, you plug in all the information you have and solve for the rate you want.

Since we want to know $\frac{dA}{dt}$, we need to know the values of $\ell$ and of $\frac{d\ell}{dt}$. The value of $\frac{d\ell}{dt}$ we already know: they told us from the beginning that it was always equal to $6$. So we just need to figure out the value of $\ell$ at the instant we are interested in.

The instant we are interested in is the instant when $A=16$. Since $A=\ell^2$, we need $16=\ell^2$; solving for $\ell$ and remembering that it is a length, so it must be nonnegative, we find that $16=\ell^2$ means that $\ell$ must be equal to $4$. Plugging in everything (the value of $\ell$, the value of $\frac{d\ell}{dt}$) into the formula we got that relates everything we get \begin{align*} \frac{dA}{dt} &=2\ell\frac{d\ell}{dt}\\ \frac{dA}{dt} &= 2(4)(6)\\ &= 48\text{ units}^2/\text{second}. \end{align*} So at the instant that the area is 16 square units, the area is changing at a rate of 48 square units per second.

-
This is incredibly difficult for me to follow with the use of dA/dt and such, also is that an antiderivative at the top? I am having trouble with the idea behind finding a relate rate between the two derivatives, I do not understand the purpose or why that works. –  user138246 May 10 '12 at 17:15
It's a related rates problem. It has to use derivatives (rates). No, there are no antiderivatives anywhere in this post. At the top I am repeating the information you were given: "Each side of the square is increasing at a rate of 6 units per second." The use of the key words "at a rate" indicates that you are being told what the rate of change (aka the derivative) of the length of the side is; that is, $\frac{d\ell}{dt}$. As to the equation $A=\ell^2$, that's the basic geometric relationship between the length of the side of a square and its area. –  Arturo Magidin May 10 '12 at 17:19
Why is the final solution 2l*l'? –  user138246 May 10 '12 at 17:25
@Jordan. The "final solution" is the rate of change of $A$ with respect to time when $A$ is 16; that is, the "final solution" is 48 square units per second, not "$2\ell\ell{}'$." The derivative of $\ell^2$ with respect to $t$ is $2\ell\ell{}'$ by the Chain Rule. You have $$\frac{d}{dt}\Bigl(\ell(t)\Bigr)^2 = 2\Bigl(\ell(t)\Bigr)^{2-1}\Bigl(\ell(t)\Bigr)'=2\ell(t)\ell{}'(t).$$ –  Arturo Magidin May 10 '12 at 17:27
Ok so I have to use implicit differentiation, related rates, power rule and all that go get the answer. Thank you. –  user138246 May 10 '12 at 17:31

$$A=s^2\implies A'=\frac{dA}{dt}=\frac{dA}{ds}\cdot\frac{ds}{dt}=2s\cdot s'=12s$$ $$A=16\implies s=4$$ So when the area is $16$, $A'=48$.

In general in such problems, you will find a linear relationship between the so-called logarithmic derivatives of the variables, i.e., $\frac{A'}{A}$ and $\frac{s'}{s}$. In this case, $A=s^2\implies \frac{A'}{A}=2\frac{s'}{s}$. If $y=cx^n$, then you get $\frac{y'}{y}=n\frac{x'}{x}$. Often such forms of equations lend themselves naturally to solving related rates problems, although admittedly it's overkill for this problem, so I don't use it here. It works particularly well when you have several variables and all constraints are polynomials in the variables, e.g. for problems involving areas and perimeters or volumes and surface areas with a possible cost function as a second constraint.

-