The length l of a rectangle is decreasing at the rate of 3cm/sec, while its width w is increasing at the rate of 3cm/sec. Find the rates of change of (1.) the area, (2.) the perimeter, (3.) the length of one diagonal at the instant when l=15 and w=6.
a = xy
da/dt = y dx/dt + x dy/dt
d^2 = x^2+y^2
Now just plug in your numbers (I used x and y instead of w and l, since the l looks like a 1)
2. The length / of a rectangle is decreasing at the rate of 3 cm/sec,
while its width w is increasing at the rate of 3 cm/sec. Find the
rates of change of (a) the area, (b) the perimeter, (c) the length of
one diagonal at the instant when I = 15 and w = 6
To find the rates of change, we need to apply the chain rule. Let's find the rates of change for each of the given quantities:
1. The Area:
The area of a rectangle is given by A = l * w. We need to find dA/dt, the rate of change of area with respect to time.
Given that dl/dt = -3 (length decreasing at the rate of 3 cm/sec) and dw/dt = 3 (width increasing at the rate of 3 cm/sec), we can substitute these values into the formula for area:
A = l * w
Differentiating both sides with respect to time (t):
dA/dt = dl/dt * w + l * dw/dt
Substituting the given values:
dA/dt = (-3) * 6 + 15 * 3
dA/dt = -18 + 45
dA/dt = 27
Therefore, the rate of change of the area is 27 cm²/sec.
2. The Perimeter:
The perimeter of a rectangle is given by P = 2(l + w). We need to find dP/dt, the rate of change of perimeter with respect to time.
Differentiating both sides of the equation P = 2(l + w) with respect to time (t):
dP/dt = 2(dl/dt + dw/dt)
Substituting the given values:
dP/dt = 2(-3 + 3)
dP/dt = 0
Therefore, the rate of change of the perimeter is 0 cm/sec.
3. The Length of One Diagonal:
The length of one diagonal of a rectangle is given by D = √(l² + w²). We need to find dD/dt, the rate of change of the length of one diagonal with respect to time.
Differentiating both sides of the equation D = √(l² + w²) with respect to time (t):
dD/dt = (1/2) * (l² + w²)^(-1/2) * (2l * dl/dt + 2w * dw/dt)
Substituting the given values:
dD/dt = (1/2) * (15² + 6²)^(-1/2) * (2 * 15 * (-3) + 2 * 6 * 3)
dD/dt = (1/2) * (225 + 36)^(-1/2) * (-90 + 36)
dD/dt = (1/2) * (261)^(-1/2) * (-54)
dD/dt = (-27/√261)
Therefore, the rate of change of the length of one diagonal is approximately -1.492 cm/sec.
To find the rates of change of the area, perimeter, and length of one diagonal, we need to derive formulas that express each quantity in terms of the length and width of the rectangle. Let's start with the area, perimeter, and diagonal formulas:
1. Area of a rectangle: A = l * w
2. Perimeter of a rectangle: P = 2 * (l + w)
3. Length of one diagonal of a rectangle: D = √(l^2 + w^2)
Next, we need to differentiate each formula with respect to time, t, and apply the given rates of change.
1. Rate of change of the area (dA/dt):
dA/dt = d(l * w)/dt
= l * dw/dt + w * dl/dt
= l * 3 + w * (-3) (Given rates of change: dl/dt = -3cm/sec, dw/dt = 3cm/sec)
2. Rate of change of the perimeter (dP/dt):
dP/dt = d(2 * (l + w))/dt
= 2 * (dl/dt + dw/dt)
= 2 * (-3 + 3) (Given rates of change: dl/dt = -3cm/sec, dw/dt = 3cm/sec)
3. Rate of change of the diagonal (dD/dt):
To find the rate of change of the diagonal, we will differentiate the formula for diagonal, D = √(l^2 + w^2), and apply the chain rule.
dD/dt = (1/2) * (l^2 + w^2)^(-1/2) * (2l * dl/dt + 2w * dw/dt)
= (l * dl/dt + w * dw/dt) * (l^2 + w^2)^(-1/2)
= (l * (-3) + w * 3) * (l^2 + w^2)^(-1/2) (Given rates of change: dl/dt = -3cm/sec, dw/dt = 3cm/sec)
Now, we can plug in the values of l and w when l = 15cm and w = 6cm to find the rates of change of each quantity:
1. Rate of change of the area (dA/dt):
dA/dt = 15 * 3 + 6 * (-3)
= 45 - 18
= 27 cm²/sec
2. Rate of change of the perimeter (dP/dt):
dP/dt = 2 * (-3 + 3)
= 2 * 0
= 0 cm/sec
3. Rate of change of the diagonal (dD/dt):
dD/dt = (15 * (-3) + 6 * 3) * (15^2 + 6^2)^(-1/2)
= (-45 + 18) * (225 + 36)^(-1/2)
= -27 * 261^(-1/2)
≈ -4.74 cm/sec
Therefore, at the instant when l = 15 and w = 6, the rates of change are:
(1.) The area is increasing at a rate of 27 cm²/sec.
(2.) The perimeter is not changing (0 cm/sec).
(3.) The length of one diagonal is decreasing at a rate of approximately 4.74 cm/sec.