Related Rates

  • Thread starter scorpa
  • Start date
  • #1
355
1
Hi Again,

I am doing a question on related rates that I have become stuck on.

The height (h) of an equilateral triangle is increasing at a rate of 3cm/min. How fast is the area changing when h is 5cm?

I know that the area of a triangle is bh/2, but after that I am stuck :redface: I tried deriving it using the chain rule so that I could substitute h and the rate of h, but I don't think that i was doing it the right way. If anyone could direct me here I would really appreciate the help.
 

Answers and Replies

  • #2
793
4
Here are some things to consider, the height "h" of an equilateral triangle is

[tex]\frac{1}{2}\sqrt{3}s[/tex]

where "s" is the length of one side.

The area of this triangle is equal to

[tex]\frac{1}{2}sh[/tex]

See any substitutions?
 
  • #3
32
0
[tex]
\frac{dA}{dt} = \frac{dh}{dt} * \frac{dA}{dh}
[/tex]
 

Related Threads on Related Rates

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
4
Views
13K
  • Last Post
Replies
4
Views
2K
Top