The altitude of a triangle is increasing at a rate of 3.0 centimeters/minute while the area of the triangle is increasing at a rate of 3.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 9.5 centimeters and the area is 94.0 square centimeters?

Question

gorv · Answer

rate of change or area= rate of change of base* rate of change of altitude

anonymous · Answer

So how would I get that? (94-3.5*9.5-3.0)?

anonymous · Answer

57.75 that's it

phi · Answer

This is a "related rates" problem in calculus.
Start with the formula for the area of a triangle
$$ A = \frac{1}{2} b\ h $$
Using the numbers given, A= 94 $ cm^2$  and h= 9.5 cm we can solve for the base
$$ b = 2A/h= 19.79 \ cm $$

Next, find the derivative with respect to time (variable "t" ). We assume all the variables A, b and h are changing.  I will use A' to show dA/dt, b' for db/dt, etc
Using the product rule we find
$$ \frac{d}{dt} \left(A = \frac{1}{2} b\ h   \right) \ 
A' =\frac{1}{2}  \left( b \ h' + h \ b'\right)  \ 2A' =  b \ h' + h \ b'
$$

Fill in the variables that we know.
altitude of a triangle is increasing at a rate of 3.0 centimeters/minute 
i.e.  h' = + 3 cm/min

area of the triangle is increasing at a rate of 3.5 square centimeters/minute. 
A' = + 3.5 cm^2 / min
 the altitude is 9.5 centimeters
h = 9.5 cm
also, from the first step, b= 19.79 cm 

we have
$$ 2 \cdot 3.5 = 19.79 \cdot 3+ 9.5 \cdot b' \
7 = 59.37 + 9.5 b' \ -52.37 / 9.5 = b' \ b'= - 5.5\ cm/min 
$$
the base is getting shorter at a rate of 5.5 cm/min