To find rate of change of area of triangle when rate of change and value of length of base and height are 3cm/min, 5cm/min and 8cm,10cm respectively.

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.