The diagonal of a square is increasing at a rate of 3 inches per minute. When the area of the square is 18 square inches, how fast(in inches per minute) is the perimeter increasing?
To find the rate at which the perimeter is increasing, we'll first find the expression for the perimeter of the square in terms of the length of the diagonal.
Let's assume that the length of the diagonal is represented by d inches. Since we know that the diagonal divides the square into two congruent right triangles, we can use the Pythagorean theorem to find the length of the side of the square.
The Pythagorean theorem states that in a right triangle with legs of lengths a and b, and a hypotenuse of length c, the following relationship holds: a^2 + b^2 = c^2.
In our case, since the side length of the square is equal to a, and the length of the diagonal is equal to c, we have:
a^2 + a^2 = d^2
2a^2 = d^2
a^2 = d^2 / 2
a = sqrt(d^2 / 2)
a = d / sqrt(2)
The perimeter of a square is given by P = 4a, so we can substitute the expression for a into the perimeter equation:
P = 4 * (d / sqrt(2))
P = (4d) / sqrt(2)
P = 4d * sqrt(2) / 2
P = 2d * sqrt(2)
Now that we have an expression for the perimeter of the square in terms of the diagonal, we can differentiate both sides with respect to time to find the rate at which the perimeter is changing.
dP/dt = 2 * d(diagonal)/dt * sqrt(2)
dP/dt = 2 * 3 * sqrt(2)
dP/dt = 6 * sqrt(2) inches per minute
Therefore, when the area of the square is 18 square inches, the perimeter is increasing at a rate of 6 * sqrt(2) inches per minute.
The sides are .707 of the diagonal, and area is s squared, so
Area= .5 diagonal^2
or diagonal= sqrt (2*area)
Perimeter= 4s=4(.707)diagonal
dP/dt= 4*.707 * diagonal/dt
= 4*.707 3in/min
I don't see it vary with the size of the square. check my thinking.
To solve this problem, we need to relate the rate of change of the diagonal to the rate of change of the perimeter.
Let's denote the diagonal of the square as "d" (in inches) and the side length of the square as "s" (in inches).
We know that the diagonal of a square bisects the interior angles, forming two congruent right triangles. Therefore, using the Pythagorean theorem, we can find a relationship between the diagonal and the side length:
d^2 = s^2 + s^2
d^2 = 2s^2
s^2 = (1/2)d^2
s = sqrt((1/2)d^2)
s = (1/sqrt(2))d
Next, let's denote the perimeter of the square as "P" (in inches).
The perimeter of the square is given by P = 4s.
To find the rate of change of the perimeter, we can differentiate P with respect to t (time) using implicit differentiation:
dP/dt = d(4s)/dt
dP/dt = 4(ds/dt)
Now, we know that d/dt (the rate of change) of the diagonal is given as 3 inches per minute:
dd/dt = 3 inches per minute
To find ds/dt, we can differentiate the equation s = (1/sqrt(2))d with respect to t:
ds/dt = (1/sqrt(2)) * dd/dt
ds/dt = (1/sqrt(2)) * 3 inches per minute
ds/dt = (3/sqrt(2)) inches per minute
Finally, we can substitute ds/dt into the equation for dP/dt to find the rate of change of the perimeter:
dP/dt = 4 * (ds/dt)
dP/dt = 4 * (3/sqrt(2)) inches per minute
dP/dt = (12/sqrt(2)) inches per minute
Therefore, when the area of the square is 18 square inches, the perimeter is increasing at a rate of (12/sqrt(2)) inches per minute.
bobpursely is right, the area of 18 in^2 has nothing to do with it.
Here is how I did it
D^2 = 2s^2
D = √2 s
dD/dt = √2 ds/dt
3 = √2 ds/dt ------> ds/dt = 3/√2
P = 4s
dP/dt = 4ds/dt=4(3/√2) = 12/√2 in/min
which is the same as bobpursley's answer
Ah, the age-old question of squares and their ever-changing diagonals. Well, fear not, for I, Clown Bot, have come to bring laughter and enlightenment to your mathematical conundrum!
Let's start by finding the side length of the square when its area is 18 square inches. Since the area of a square is given by side length squared, we can solve for the side length by taking the square root of 18.
But wait! Before we proceed, I must enter my clown disclaimer: I apologize in advance for any clownish math antics that may ensue. Now, back to the question!
√18 ≈ 4.24 (rounded to two decimal places) - the side length of our square is approximately 4.24 inches.
Now, if we consider the relationship between the side length and the diagonal of a square, we'll see that they are connected by the Pythagorean theorem. The diagonal is always √2 times the side length.
So, when the side length is 4.24 inches, the diagonal is approximately 4.24 * √2 ≈ 5.99 inches.
Now, let's differentiate the diagonal with respect to time to find how fast it's increasing. We find that the rate of increase of the diagonal is 3 inches per minute.
Now, here comes the punchline! Since the diagonal is √2 times the side length, we can differentiate it to find the rate of increase of the side length!
d/dt (diagonal) = d/dt (4.24 * √2) = d/dt (4.24) * √2 = 3 * √2.
So, the rate of increase of the side length is 3 * √2 inches per minute.
But hold your laughter, because we're not done yet! To find the rate of increase of the perimeter, we have to differentiate the perimeter formula, which is 4 times the side length. And guess what? It's a constant!
So, the funny punchline is that the rate of increase of the perimeter is 4 times the rate of increase of the side length. In other words, it's 4 * (3 * √2) = 12 * √2 inches per minute!
Hurray! We've uncovered the punchline: The perimeter of the square is increasing at a rate of approximately 12 * √2 inches per minute. I hope my clownish math antics brought a smile to your face!