# Could I get some help finding how fast a triangle's area is changing?

• March 7th 2009, 06:38 PM
Meeklo Braca
Could I get some help finding how fast a triangle's area is changing?
The question is:

The height h of an equilateral triangle is increasing at a rate of 3cm/min. How fast is the area changing when h is 5 cm? Give the answer to 2 decimal places.

How can I figure this question out?

Thank You!
• March 7th 2009, 11:52 PM
earboth
Quote:

Originally Posted by Meeklo Braca
The question is:

The height h of an equilateral triangle is increasing at a rate of 3cm/min. How fast is the area changing when h is 5 cm? Give the answer to 2 decimal places.

How can I figure this question out?

Thank You!

1. The area of a triangle is calculated by

$a = \dfrac12 \cdot s \cdot h$

2. In an equilateral triangle the sides are equal and the height is calculated by:

$h^2+\left(\frac12 s\right)^2 = s^2~\implies~h^2 = \dfrac34 s^2~\implies~s = \dfrac23 \sqrt{3} \cdot h$

3. Plug in this term instaed of s into the equation of the area:

$a(h)=\dfrac13 \sqrt{3} \cdot h^2$

4. Calculate the first derivation of a to get the speed of change:

$a'(h) = \dfrac23 \sqrt{3} \cdot h$

Now calculate a'(5):

$a'(5) = \dfrac23 \sqrt{3} \cdot 5 ~\approx 5.77 \ square\ units$
• March 9th 2009, 04:37 PM
Meeklo Braca
Isnt the derivative equation

da/dt=(1/3)sqrt3h * dh/dt

Isnt your equation missing the dh/dt with a final answer being 8.66?
• March 9th 2009, 08:44 PM
mr fantastic
Quote:

Originally Posted by Meeklo Braca
Isnt the derivative equation

da/dt=(1/3)sqrt3h * dh/dt

Isnt your equation missing the dh/dt with a final answer being 8.66?

earboth has given you $\frac{da}{dh}$. To get the final answer you must multiply this by $\frac{dh}{dt} = 3$ cm/min.