How Fast Is The Area Of The Triangle Changing
- #1
The question is:
The peak h of an equilateral triangle is increasing at a charge per unit of 3cm/min. How fast is the area changing when h is v cm? Give the answer to 2 decimal places.
How tin can I figure this question out?
Thanks!
- January 2006
- 5,854
- 2,554
- Germany
- #ii
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 v cm? Give the respond to 2 decimal places.
How can I effigy this question out?
Thank Y'all!
1. The expanse of a triangle is calculated by
\(\displaystyle a = \dfrac12 \cdot s \cdot h\)
2. In an equilateral triangle the sides are equal and the tiptop is calculated by:
\(\displaystyle h^two+\left(\frac12 s\right)^2 = s^ii~\implies~h^ii = \dfrac34 s^2~\implies~southward = \dfrac23 \sqrt{iii} \cdot h\)
3. Plug in this term instaed of southward into the equation of the expanse:
\(\displaystyle a(h)=\dfrac13 \sqrt{3} \cdot h^2\)
4. Calculate the outset derivation of a to get the speed of change:
\(\displaystyle a'(h) = \dfrac23 \sqrt{3} \cdot h\)
Now calculate a'(5):
\(\displaystyle a'(5) = \dfrac23 \sqrt{three} \cdot 5 ~\approx 5.77 \ square\ units\)
- #3
Isnt the derivative equation
da/dt=(1/three)sqrt3h * dh/dt
Isnt your equation missing the dh/dt with a last answer being 8.66?
- Dec 2007
- 16,947
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- #four
Isnt the derivative equation
da/dt=(1/3)sqrt3h * dh/dt
Isnt your equation missing the dh/dt with a last reply being eight.66?
earboth has given y'all \(\displaystyle \frac{da}{dh}\). To get the final reply you lot must multiply this past \(\displaystyle \frac{dh}{dt} = 3\) cm/min.
Terminal edited:
Source: https://mathhelpforum.com/threads/could-i-get-some-help-finding-how-fast-a-triangles-area-is-changing.77422/
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