Serious calculus help needed.?
The sides of an equilateral triangle are increasing at the rate of 27 in/sec. How fast is the triangle's area increasing when the sides of the triangle are each 18 inches long?
- NorthstarLv 71 decade agoFavorite Answer
It's unclear whether each side is increasing at the rate of 27 in/sec or whether the perimeter is increasing at that rate. I am making the assumption that each side is increasing at that rate.
The area of an equilateral triangle with side s is:
A = (√3/4)s^2
ds/dt = 27
dA/ds = (√3/2)s
dA/dt = (dA/ds)(ds/dt) = (√3/2)s*27 = (27√3/2)s
= (27√3/2)(18) = 243√3 = 420.9 ft^2/sec
- PuggyLv 71 decade ago
Let s = the side of the triangle. We're given that
ds/dt = 27
We WANT to know what dA/dt is, where A is the area of the triangle.
In order to solve this problem, we must relate A and s.
We know that A = base times height, and our base is s but we need to know what height is.
If you bisect an equilateral triangle, you'll notice that it forms two right angle triangles, with hypotenuse s and base s/2. That makes the height solveable by the Pythagoras Theorem.
h^2 + (s/2)^2 = s^2, so
h^2 = s^2 - (s/2)^2
h^2 = s^2 - (s^2)/4, or
h^2 = (3/4) s^2. Taking the square root of both sides, we get
h = [sqrt(3)/2]s
Since A = base times height, and base = s, height = [sqrt(3)/2]s:
A = s[sqrt(3)/2]s, or
A = [sqrt(3)/2]s^2.
When solving related rates problem, we ALWAYS differentiate with respect to t. We have to solve this using implicit differentiation.
dA/dt = 2[sqrt(3)/2]s (ds/dt)
However, we know that ds/dt is equal to, which is given to be 27, so
dA/dt = 2[sqrt(3)/2]s , or
dA/dt = [27sqrt(3)] s
In every related rates question, we have our "when" statement. It would be incorrect to plug in our numeric value (s = 18) early, and must be plugged in only AFTER the implicit differentiation and substitution of rates.
WHEN THE SIDES OF THE TRIANGLE ARE EACH 18 INCHES LONG...
Now, we plug in s = 18.
dA/dt = [27sqrt(3)] 
dA/dt = 486 sqrt(3)
Now, your concluding statement would be:
The triangle's area is increasing at a rate of 486 sqrt(3) square inches per second.
- feanorLv 71 decade ago
Area of equalateral triangle = s^2 * sqrt(3)/4
dA/dt = s*sqrt(3)/2 ds/dt = 18 * sqrt(3)/2 * 27 = 243 * sqrt(3)
- pereyraLv 44 years ago
2(x^2+y^2)^2=25(x^2–y^2) differentiating implicitly 4(x^2 + y^2)(2x + 2yy ') = 25(2x - 2yy') divide by making use of two 4(x^2 + y^2) (x + yy') = 25(x - yy') 4(x^3 + xy^2 + x^2yy' + y^3y') = 25x - 25yy' y' (4x^2 y + 4y^3 + 25y) + x(4x^2 + 4y^2 - 25) = 0 y ' = x(25 - 4x^2 - 4y^2) /(4x^2 + 4y^3 + 25y) substitute x = 3 and y =a million y ' = 3(25 - 36 - 4) / (36 + 4 + 25) = -40 5 / sixty 5 = -9/13