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.

calculusrelated-rates

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?

Best Answer

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.

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