Assume the dimensions of a rectangle are
continuously changing in a way so that the area, A, of the rectangle
remains constant. If the base of the rectangle is increasing at a rate
of 3 inches per second, at what rate is the height of the rectangle
changing at the instant when the rectangle is actually a square? Is
the height increasing or decreasing at this instant? Include units.
Show all work.
let the base be x
let the height be y
then
A = xy
dA/dt = x dy/dt + y dx/dt
but dA/dt = 0
x dy/dt = -y dx/dt
given:
dx/dt = 3,
find : dy/dt when x = y
x dy/xt = -x(3)
dy/dt = -3
So at the moment when it turns into a square, the height is decreasing at 3 inches/second
Thanks!
To find the rate at which the height of the rectangle is changing when the rectangle is a square, we need to use the concept of related rates.
Let's denote the base of the rectangle as x inches and the height as y inches. We are given that the area remains constant. Thus, we have the equation:
A = xy.
Differentiating both sides of the equation with respect to time (t), we get:
dA/dt = (d/dt) (xy).
By the product rule of differentiation, the equation becomes:
dA/dt = x(dy/dt) + y(dx/dt).
Since the area remains constant, dA/dt is zero. We are also given that dx/dt = 3 inches/second.
Now, when the rectangle is a square, the base and height are equal. So, x = y.
Substituting the given values into the equation, we get:
0 = x(dy/dt) + x(3).
Now, when the rectangle is a square, x = y, so we can rewrite the equation as:
0 = y(dy/dt) + y(3).
Simplifying the equation, we have:
0 = y(dy/dt + 3).
To find the rate at which the height is changing, we need to solve for dy/dt. From the equation, we can see that dy/dt = -3 inches/second.
Since dy/dt is negative, the height of the rectangle is decreasing at this instant.
So, the rate at which the height of the rectangle is changing when the rectangle is a square is -3 inches/second, and the height is decreasing.