Question 1084628
<pre><b><font size=4>
>>the altitude (height h) of a triangle is increasing 
at a rate of 1 cm/min<< 

{{{dh/dt}}}{{{""=""}}}{{{matrix(1,2,1,cm/(min))}}}

>>while the area A of the triangle is increasing at a 
rate of 2cm^2 /min.<< 

{{{dA/dt}}}{{{""=""}}}{{{matrix(1,2,2,cm^2/(min))}}}

>>at what rate is the base of the triangle changing...<<

{{{A}}}{{{""=""}}}{{{expr(1/2)*b*h}}}

{{{dA/dt}}}{{{""=""}}}{{{expr(1/2)(b*expr(dh/dt)+h*expr(db/dt))}}} 

Substitute {{{dA/dt}}}{{{""=""}}}{{{2}}} and {{{dh/ht}}}{{{""=""}}}{{{1}}}

{{{2}}}{{{""=""}}}{{{expr(1/2)(b*(1)+h*expr(db/dt))}}}

Multiply both side by 2

{{{4}}}{{{""=""}}}{{{2*expr(1/2)(b+h*expr(db/dt))}}}

{{{4}}}{{{""=""}}}{{{b+h*expr(db/dt)}}}

We will need b and h to solve for {{{db/dt}}}.

>>when the altitude (height h) is 10cm and the area is 100cm^2?<<

We go back to the area formula:

{{{A}}}{{{""=""}}}{{{expr(1/2)*b*h}}}
{{{100}}}{{{""=""}}}{{{expr(1/2)*b*10}}}
{{{100}}}{{{""=""}}}{{{5*b}}}
{{{20}}}{{{""=""}}}{{{b}}}

Substitute {{{h}}}{{{""=""}}}{{{10}}} and {{{b}}}{{{""=""}}}{{{20}}} in

{{{4}}}{{{""=""}}}{{{b+h*expr(db/dt)}}}
{{{4}}}{{{""=""}}}{{{20+10*expr(db/dt)}}}
{{{-16}}}{{{""=""}}}{{{10*expr(db/dt)}}}
{{{(-16)/10}}}{{{""=""}}}{{{db/dt}}}
{{{-1.6}}}{{{""=""}}}{{{db/dt}}}

The negative sign means the base is DEcreasing
at the rate of -1.6 cm/min at that instant.

Edwin</pre>