Somewhat Intense Calc Problem?

Question

The length L of a rectangle is decreasing at the rate of 2 cm/sec while the width w is increasing at the rate of 2 cm/sec. When L=12cm and W=5cm, find the rates of change of
a. the area
b. the perimeter
c. length of a diagonal of the rectangle

This is a related rate equation

How did you do that?

Answer 1

Not that difficult, if you are good at calculus.

a. Area increases at rate of 14 cm^2/sec
b. Perimeter stays constant
c. Diagonal decreases at 14/13 cm/sec

Area A = L*W
Rate of length change = dL/dt
Rate of width change = dW/dt

Rate of Area change = dA/dt = d(L*W)/dt = L*(dW/dt) + W*(dL/dt)
= 12*(+2) + 5*(-2) = 24 - 10 = 14

Perimeter = 2*(L+W)

Rate of Perimeter change = d{ 2*(L+W) }/dt = 2*( dL/dt + dW/dt )
2*{ (-2) + (+2) } = 0

Length of diagonal = D = sqrt( L^2 + W^2 )

D^2 = L^2 + W^2
differentiate both sides;
2D(dD/dt) = 2L(dL/dt) + 2W(dW/dt)
D(dD/dt) = L(dL/dt) + W(dW/dt)
dD/dt = (1/D)*{ 12*(-2) + 5*(+2) } = -14/D = -14/sqrt(12^2 + 5^2)
= -14/sqrt(169) = (-14/13)

Answer 2

a.
A = LW
dA/dt = dL/dt * W + L (dW/dt)
dA/dt = (-2) (5) + (12)(2)
dA/dt = -10 + 24
dA/dt = 14 cm^2/sec

b.
P = 2L + 2W
dP/dt = 2(dL/dt) + 2(dW/dt)
dP/dt = 2(-2) + 2(2)
dP/dt = 0

c.
D = (L^2 + W^2)^(1/2)
dD/dt = 1/2(L^2 + W^2)^(-1/2) * (2L(dL/dt) + 2W(dW/dt))
dD/dt = 1/2(144 + 25)^(-1/2) *(2(12)(-2) + 2(5)(2))
dD/dt = 1/2(169)^(-1/2) * (-48 + 20)
dD/dt = 1/(2*13) * (-28)
dD/dt = -14 / 13
dD/dt = -1.077 cm/sec

Answer 3

http://img175.imageshack.us/img175/41/untitledlt8.jpg