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 l = 15 and w = 6.
that's better. See what you can do with (b) and (c), from the steps I used in your previous post.
Post your work if you get stuck.
To find the rates of change of the area, perimeter, and length of one diagonal of the rectangle, we can use the chain rule of differentiation. The formulas we need are:
(a) Rate of change of the area:
A = l * w
where A is the area, l is the length, and w is the width.
(b) Rate of change of the perimeter:
P = 2l + 2w
where P is the perimeter, l is the length, and w is the width.
(c) Length of the diagonal:
d = sqrt(l^2 + w^2)
where d is the length of the diagonal, l is the length, and w is the width.
Given that the length is decreasing at a rate of 3 cm/sec and the width is increasing at a rate of 3 cm/sec, we can differentiate these equations with respect to time (t) to find the rates of change.
Let's find the rates of change of the area, perimeter, and length of one diagonal when l = 15 and w = 6.
(a) Rate of change of the area:
Differentiating the equation A = l * w with respect to t, we get:
dA/dt = dl/dt * w + l * dw/dt
Substituting the given rates of change, we have:
dA/dt = -3 * 6 + 15 * 3
dA/dt = -18 + 45
dA/dt = 27 cm²/sec
(b) Rate of change of the perimeter:
Differentiating the equation P = 2l + 2w with respect to t, we get:
dP/dt = 2 * dl/dt + 2 * dw/dt
Substituting the given rates of change, we have:
dP/dt = 2 * (-3) + 2 * 3
dP/dt = -6 + 6
dP/dt = 0 cm/sec
(c) Rate of change of the length of one diagonal:
Differentiating the equation d = sqrt(l^2 + w^2) with respect to t, we get:
dd/dt = (1/2) * (l^2 + w^2)^(-1/2) * (2l * dl/dt + 2w * dw/dt)
Substituting the given rates of change, we have:
dd/dt = (1/2) * (15^2 + 6^2)^(-1/2) * (2 * 15 * (-3) + 2 * 6 * 3)
dd/dt = (225 + 36)^(-1/2) * (-90 + 36)
dd/dt = 261^(-1/2) * (-54)
dd/dt = -54/√261 cm/sec
Therefore, at the instant when l = 15 cm and w = 6 cm, the rates of change are:
(a) The area is increasing at 27 cm²/sec.
(b) The perimeter is not changing (0 cm/sec).
(c) The length of one diagonal is decreasing at approximately -4.80 cm/sec.
To find the rates of change of (a) the area, (b) the perimeter, and (c) the length of one diagonal, we need to use the formulas related to these quantities.
(a) The formula for the area of a rectangle is A = length × width. We need to find the rate of change of the area, which is dA/dt. The rates of change of length (dl/dt) and width (dw/dt) are given.
Using the chain rule, we have:
dA/dt = ∂A/∂l × dl/dt + ∂A/∂w × dw/dt
Since ∂A/∂l = width and ∂A/∂w = length, we substitute the known values:
dA/dt = width × (-3) + length × 3
dA/dt = 6 × (-3) + 15 × 3
dA/dt = -18 + 45
dA/dt = 27 cm²/sec
Therefore, the rate of change of the area is 27 cm²/sec.
(b) The formula for the perimeter of a rectangle is P = 2(length + width). We need to find the rate of change of the perimeter, which is dP/dt.
Using the chain rule, we have:
dP/dt = 2 × (dl/dt + dw/dt)
Substituting the known values:
dP/dt = 2 × (-3 + 3)
dP/dt = 0 cm/sec
Therefore, the rate of change of the perimeter is 0 cm/sec.
(c) The formula for the length of the diagonal of a rectangle is given by the Pythagorean theorem:
d² = l² + w²
To find the rate of change of the length of the diagonal, which is dd/dt, we need to differentiate this equation implicitly with respect to time (t).
2d × dd/dt = 2l × dl/dt + 2w × dw/dt
Substituting the known values when l = 15 and w = 6:
2 × (dd/dt) = 2 × 15 × (-3) + 2 × 6 × 3
2 × (dd/dt) = -90 + 36
2 × (dd/dt) = -54
dd/dt = -54/2
dd/dt = -27 cm/sec
Therefore, the rate of change of the length of one diagonal is -27 cm/sec.