Related Rates problem!!
Favorites|Homepage
Subscriptions | sitemap
HOME > > Related Rates problem!!

Related Rates problem!!

[From: ] [author: ] [Date: 12-02-12] [Hit: ]
Therefore, in an equilateral triangle with height h, *half* of the base is h/sqrt(3).So the area is A = h(h/sqrt(3)) = h^2 / sqrt(3).dA/dt = [2h/sqrt(3)](dh/dt).Substituting dh/dt = 3 and h = 5,......
The height h of an equilateral triangle is increasing at a rate of 3 cm/min. How fast is the area changing when h is 5 cm? Give your answer correct to two decimal places.

-
Step 1: List all your given information...
dh/dt = 3 cm/min
h = 5 cm
dA/dt = ?

Step 2: Determine a relationship between your starting and ending variables
A = 1/2 * base * height

Since it is an equilateral triangle, you can determine the base in terms of h.
base = 2sqrt(3) / 3 * h
So,
A = sqrt(3)/3 * h^2

Finally, take the derivative
dA/dt = sqrt(3)/3 * (2h) * dh/dt
Step 3: Solve for answer
dA/dt = sqrt(3)/3 * 2 * (5) * (3)

dA/dt = 10 sqrt(3) cm^2

-
The altitude of an equilateral triangle divides the triangle into two 30-60-90 triangles.
In a 30-60-90 triangle, the leg opposite the 30-degree angle is 1/sqrt(3) times the leg opposite the 60-degree angle.
Therefore, in an equilateral triangle with height h, *half* of the base is h/sqrt(3).
So the area is A = h(h/sqrt(3)) = h^2 / sqrt(3).

Differentiating with respect to t gives
dA/dt = [2h/sqrt(3)](dh/dt).

Substituting dh/dt = 3 and h = 5, we find that the area increases at a rate of
dA/dt = [2*5/sqrt(3)](3) = 10sqrt(3) or about 17.32 cm^2/min.

Lord bless you today!

-
You are searching for dA/dt when h is 5 cm. IGNORE the 5 cm until the end. Your job is to solve while incorporating 3 cm/min, or your dh/dt.

dA/dt = dh/dt x dA/dh

You simply have to create an equation for Area in terms of h only. Because the triangle is equilateral, there is a way to cancel out the b in the equation A=0.5bh. Keep in mind that you may have to draw a picture and create extra variables in the process. (I'd be willing to bet that there will be trigonometry involved.)

Once you find this equation, solve it for when h=5, then multiply the answer by 3, and you will have the exact answer.

Sorry I didn't write any math down, but I just organized the set up because it's all I can do right now.

Hope this helps! ;)
1
keywords: Rates,problem,Related,Related Rates problem!!
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .