# related rate problem

two fast ants start traveling from the same point in the sand. one ant heads north at a rate of 2cm/s and the other Ant travels east @ a rate of 3cm/s. the trails the ant leave form the sides of a right triangle. how fast is the area of the triangle increasing after 5 seconds.

i took the following approach i know the area of a triangle is $$\frac{1}{2}bh$$ and im given that the base of my triangle is changing at a rate of 3 cm's and my height if changing at a rate of 2 cm's therefore i said. $$\frac{dh}{dt} = 2$$ $$\frac{db}{dt} = 3$$

so i take the derivative of my triangle formula and i end up with
$$A \frac{da}{dt}=\frac{1}{2} \frac{db}{dt} \frac{dh}{dt}$$ substituting the values in im solving for $\frac{da}{dt}$ $$5\frac{da}{dt}=\frac{1}{2}*2*3$$ $$\frac{da}{dt}=\frac{3}{5}cm/s$$ did i approach this correctly? i would love know if i did this correctly but im not 100% sure. I like to get someones second opinion on it. Thanks Miguel

• Be careful, the derivative of a product is not the product of the derivatives : $(uv)'=u'v+uv'\neq u'v'$ ;) – Dolma Apr 15 '13 at 20:54
• Since the area of the triangle $a = \frac{1}{2} bh$ involves a product, don't you want to use the Product Rule when you implicitly differentiate $a$ ? Also, what is $A$ and why have you inserted 5 (seconds) there? – colormegone Apr 15 '13 at 20:56
• also your units should be $cm^2/sec$. This is another hint for the product rule. If you evaluate your units they don't match. – Ross Millikan Apr 15 '13 at 21:07

You need to use the product rule. From $a=\frac 12bh$ you get $a'=\frac 12 b'h+\frac 12 bh'$

Added: See if the below figure helps. I drew a rectangle because it seemed clearer. At any time, the rectangle area is $a=bh$. If $b$ and $h$ both increase by $db$ and $dh$ we add the two long rectangles $db \times h$ and $dh \times b$ The small $dh \times db$ is negligible.

• so should it be $$\frac{dh}{dt}+\frac{db}{dt}$$? iw that where i went wrong? – Miguel Apr 15 '13 at 20:56
• @Miguel: no, it should be $\frac 12(b\frac {dh}{dt}+h\frac {db}{dt})$. This should be in your text. – Ross Millikan Apr 15 '13 at 20:58
• thank you i get it now. i see where i went wrong. – Miguel Apr 16 '13 at 13:00

You can use what you just said, you know the derivative of both sides: $\frac{dh}{dt}=2$ and $\frac{db}{dt}=3$.

This tells you that $h=2t+h_0$ and $b=3t+b_0$.

Here $h_0=b_0=0$ so $h=2t$ and $b=3t$. Thus $A=\frac{1}{2}(2t)(3t)=3t^2$.

You can now say that $\frac{dA}{dt}=6t$

Now you know how fast it goes after 5 seconds ($t=5$)

• Just something you should check which could help you spot a mistake: if you're working with an area and both sides are evolving at fixed rate, then it's evolving at a squared rate. Which means the rate increases as time goes by. If it's a volume then it's some cubed rate, which means the rate at which the rate evolves is increasing. And so on ... – Dolma Apr 15 '13 at 21:07
• thank you this is very helpful. – Miguel Apr 16 '13 at 12:59