A website features a rectangular display with the dimensions of the rectangle changing continuously. At what rate is the height of the rectangle changing when it (the height) is 3 cm and the diagonal of the rectangle is 5 cm?

Given that the area of the rectangle is increasing at 3/4 cm^2 per second and the diagonal of the rectangle is increasing at 1/3 cm per second.

dy/dt = - 0.0513 cm/s

Step-by-step explanation:

Given

dy/dt = ?

y = 3 cm (the height of the rectangle)

D = 5 cm (the diagonal of the rectangle)

dA/dt = 3/4 cm²/s

dD/dt = 1/3 cm/s

We can apply the formula

A = x*y ⇒ x = A/y

where x is the base and A is the area.

If we use Pythagoras' theorem

x² + y² = D² (i)

⇒ (A/y) ² + y² = D²

we apply

((A/y) ²) ' + (y²) ' = (D²) '

2 * (A/y) * (((dA/dt) * y - A * (dy/dt)) / y²) + 2*y * (dy/dt) = 2*D * (dD/dt)

⇒ (A/y) * (((dA/dt) * y - A * (dy/dt)) / y²) + y * (dy/dt) = D * (dD/dt)

⇒ (dy/dt) * (y - (A²/y³)) = D * (dD/dt) - (A/y²) * (dA/dt)

⇒ dy/dt = (D * (dD/dt) - (A/y²) * (dA/dt)) / (y - (A²/y³)) (ii)

from eq. (i) we have

x² + (3 cm) ² = (5 cm) ² ⇒ x = 4 cm

we obtain A:

A = x*y ⇒ A = 4 cm * 3 cm

⇒ A = 12 cm²

Finally, we use eq. (ii)

dy/dt = (5 cm * (1/3 cm/s) - (12 cm² / (3 cm) ²) * (3/4 cm²/s)) / (3 cm - ((12 cm²) ² / (3 cm) ³))

⇒ dy/dt = - 0.0513 cm/s