Question

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.

Answer 1

dy/dt = - 0.0513 cm/s

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