The side of a square are increasing at 2 inches per second. How fast is the diagonal increasing when the side is 3 inches?

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Diagonal --- d

side ---x

d^2 = x^2 + x^2
d = √2x
dd/dt = √2 dx/dt
at any length of x
dd/dt = √2(2)
= 2√2 inches per second

thank you

To find how fast the diagonal of the square is increasing, we need to use the concept of related rates. Let's denote the side of the square as "s" and the diagonal as "d".

Given that the side of the square is increasing at a rate of 2 inches per second, we can express this as ds/dt = 2, where ds/dt represents the rate of change of the side with respect to time.

We are asked to find how fast the diagonal is increasing when the side is 3 inches. Therefore, we need to determine the rate of change of the diagonal with respect to time, dd/dt, when s = 3 inches.

We can use the Pythagorean theorem to relate the side length and the diagonal of a square:

d² = s² + s² = 2s²

Differentiating both sides of the equation with respect to time (t), we get:

2d * dd/dt = 4s * ds/dt

Since we have values for ds/dt, s, and d, we can substitute them in:

2d * dd/dt = 4(3)(2)

Now we can solve for dd/dt:

2d * dd/dt = 24

dd/dt = 24 / (2d)

To find the value of dd/dt when s = 3 inches, we need to find the value of d. In a square, the diagonal is equal to the side multiplied by sqrt(2), so when the side is 3 inches, the diagonal is:

d = 3 * sqrt(2)

Substituting this value into the expression for dd/dt:

dd/dt = 24 / (2 * 3 * sqrt(2))

Simplifying further:

dd/dt = 4 / sqrt(2)

Rationalizing the denominator:

dd/dt = 4 / (sqrt(2) * sqrt(2))

dd/dt = 4 / 2

dd/dt = 2 inches per second

Therefore, the diagonal of the square is increasing at a rate of 2 inches per second when the side is 3 inches.