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# 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?

### 4 Answers

- 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 [27], 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)] [18]

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