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

1. ## 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!

2. 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

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

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

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

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

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

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

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

Now calculate a'(5):

$\displaystyle a'(5) = \dfrac23 \sqrt{3} \cdot 5 ~\approx 5.77 \ square\ units$

3. 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?

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

,

,

,

,

### the side of an equilateral triangle is increasing at the rate of √3 cm/sec.how fast is the area increasing when its side is 5 cm.

Click on a term to search for related topics.