# 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?

Relevance

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

• 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.

• 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)

• 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