I am trying understand very simple related rates problem (area of triangle on youtube):

The base of a right triangle is increasing at 3cm/min while the height of the triangle is increasing at a rate of 5cm/min. How fast is the area of the triangle changing when the base and height are 8cm and 10cm long respectively.

Base of the triangle is defined as $b = 8cm$, height as $h = 10cm$. Base increases 3 cm per min so

$\frac{db}{dt} = 3 cm$

Height increases 5cm per min so

$\frac{dh}{dt} = 5cm$

Question is about finding how fast is area of triangle changes. Triangle area is of course defined as

$A = \frac{1}{2}bh$

and its derivative is

$\frac{dA}{dt} = \frac{1}{2}\frac{db}{dt}h + \frac{1}{2}\frac{dh}{dt}b$

By substituting numbers we know we get

$\frac{dA}{dt} = \frac{1}{2}*3*10 + \frac{1}{2}*5*8 = 35$

So my understanding is that area of triangle is changing $35 cm^2$ per minute.

So far everything is clear but I am confused when I tried to compute real area after base and heigh is increased. With $b = 8cm$ and $h = 10cm$ area is $40 cm^2$. After one minute if area is increasing by $35cm^2$ it should be $75 cm^2$. But when I substitute new base and height I got different number:

$A = \frac{1}{2}*11*15 = 82.5$

Here I am confused as I would expect area to match original area + rate of change. What exactly that $35 cm^2$ means and why those numbers does not match?

  • 4
    $\begingroup$ I think expressing your rates of changes as just units of length and area, and not length or area per unit time, is one problem point. $db/dt$ is 3 cm/min, not just 3 cm, for instance. The new area is not just going to blindly be the original plus the rate of change, for the same reason you can't even add 4 m and 3 m/s. $\endgroup$ Oct 2 at 14:26
  • 2
    $\begingroup$ You calculated the instantaneous rate of change, which means the area will be changing at the rate of $35 cm^2/min$ one minute after the start. It's not the same as change in total area. $\endgroup$
    – Vasili
    Oct 2 at 14:32
  • 2
    $\begingroup$ while taking dA/dt you took h=10 and b=8 which is not always true. values of h and b are changing with time so dA/dt will also depend on time and you will have to integrate to fond the area at any time. $\endgroup$
    – Scilent144
    Oct 3 at 4:54

3 Answers 3


As Vasili points out in the comments, the area's rate of change is not constant. In order to find the area after a given amount of time, you need to integrate the rate of change over that time.

In your case, you have: $$ \frac{dA}{dt}(t) = \frac{1}{2}[3h(t) + 5b(t)], $$ with the area after $1$ minute being: $$ A(1) = A(0) + \int_0^1\frac{dA}{dt}(t)dt. $$ Since the rates of change of base and height are constant, you can simply write them as: $$ h(t) = 10 + 5t,\hspace{5mm} b(t) = 8 + 3t, $$ meaning that you can compute the area after $1$ minute as: $$ A(1) = 40 + \frac{1}{2}\int_0^1[3(10+5t)+5(8+3t)]dt = 40 + \frac{1}{2}\left(70 + 30\int_0^1t dt\right) = 40 + 42.5 = 82.5, $$ which is exactly the result you have.

  • $\begingroup$ What if the rates of change of base and height weren't constant? Then would we have degree-two equations for them? $\endgroup$
    – user724085
    Oct 2 at 23:00
  • $\begingroup$ @SnackExchange If the rates of change were linear with time, then yes, we would have a degree-two polynomial for base and height. In general, if the rates of change were more complicated function of time, we would need to integrate them to find explicit expressions for $h(t)$ and $b(t)$. $\endgroup$ Oct 3 at 9:11

Maybe a (kind of) simpler answer will help too.

So, if you have $$\frac{db}{dt}=3\ \mathrm{and}\ \frac{dh}{dt}=5,$$ the base and height (as functions of time) will be $$b(t)=3t+C_1\ \mathrm{and}\ h(t)=5t+C_2.$$

To find the actual values for $C_1$ and $C_2$, plug the initial conditions in: $b(0)=8$, $h(0)=10$ (so you'll get $C_1=8$ and $C_2=10$).

The area (again, as a function of time) will be $$A(t)=\frac12 b(t)h(t)=\frac12(3t+8)(5t+10)=7.5t^2+35t+40,$$

and the derivative (speed of change) is $$\frac{dA}{dt}=15t+35$$

So, at $t=0$ the area (which is 40 cm² at the moment) is indeed changing at 35 cm²/min, but after a minute, at $t=1$, the change has been accelerated to 50 cm²/min (and the area has indeed grown to 82.5 cm²). That's because the speed of change also depends on the time ($\frac{dA}{dt}$ is not constant, but depends on $t$ itself).

Consider an analogy with a car: if you're now driving a car at, say, 80 km/h, that does not necessarily mean that you'll drive 80 km during the next hour. It may be e.g. 50 or 130 km, and it will depend on how fast you'll be driving for the whole hour, not a single point in time.


The problem with your approach is that you are thinking of derivatives as a discrete phenomenon. That is the area of the values of base length and height are not going to increase only increase after the one minute span but it is increasing throughout that time interval. This is because it is an continuous property. So when you say that the rate of change at the moment is so and so value and thus it should increase by this much amount, it is not correct. Because the values of your height and base length is going to increase too in that one minute and thus increasing your rate of change of area. Thus the increase in the area would be more than you expect at the beginning of that minute, which is the case in this scenario. I hope i was able to explain this properly. Please don’t mind my grammar and informality. Any suggestions would be greatly appreciated.


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