The length ℓ, width w, and height h of a box change with time. At a certain instant the dimensions are ℓ = 7 m and w = h = 2 m, and ℓ and w are increasing at a rate of 3 m/s while h is decreasing at a rate of 1 m/s. At that instant find the rates at which the following quantities are changing. The length of a diagonal. (Round your answer to two decimal places.)
the main diagonal z can be found using
z^2 = ℓ^2 + w^2 + h^2
z dz/dt = ℓ dℓ/dt + w dw/dt h dh/dt
So, plug in your numbers and solve for dz/dt
If it's some other diagonal you want, then solve using only two dimensions.
d^2 = ℓ^2 + w^2 + h^2
differentiating ... 2 d dd/dt = 2 ℓ dℓ/dt + 2 w dw/dt + 2 h dh/dt
divide by 2 and substitute ... √57 dd/dt = (7 * 3) + (2 * 3) + (2 * -1)
solve for dd/dt
To find the rate at which the length of the diagonal is changing, we can use the chain rule of differentiation.
The length of the diagonal (D) of the box can be calculated using the Pythagorean theorem:
D = √(ℓ^2 + w^2 + h^2)
We are given that ℓ is increasing at a rate of 3 m/s, w is increasing at a rate of 3 m/s, and h is decreasing at a rate of 1 m/s.
Differentiating both sides of the equation with respect to time (t), we get:
dD/dt = (1/2)(ℓ^2 + w^2 + h^2)^(-1/2)(2ℓ dℓ/dt + 2w dw/dt + 2h dh/dt)
Substituting the given values:
dD/dt = (1/2)(7^2 + 2^2 + 2^2)^(-1/2)(2*7*3 + 2*2*3 + 2*2*(-1))
Simplifying the equation:
dD/dt = (1/2)(57)^(-1/2)(42 + 12 - 8)
dD/dt = (1/2)(57)^(-1/2)(46)
Calculating the value:
dD/dt ≈ 0.487 m/s
Therefore, at that instant, the length of the diagonal is changing at a rate of approximately 0.487 m/s.
To find the rate at which the length of the diagonal of the box is changing, we need to use the chain rule from calculus.
The length of the diagonal of the box can be found using the Pythagorean theorem. Let's call the length of the diagonal D.
D = √(ℓ^2 + w^2 + h^2)
Now, we need to find the derivative of D with respect to time. Let's denote this as dD/dt.
Since ℓ and w are increasing and h is decreasing with respect to time, we can express their rates of change as follows:
dℓ/dt = 3 m/s (positive because it is increasing)
dw/dt = 3 m/s (positive because it is increasing)
dh/dt = -1 m/s (negative because it is decreasing)
Using the chain rule, we can differentiate D with respect to time:
dD/dt = (dD/dℓ) * (dℓ/dt) + (dD/dw) * (dw/dt) + (dD/dh) * (dh/dt)
To find the partial derivatives, we differentiate D with respect to each variable:
dD/dℓ = (1/2)*(ℓ^2 + w^2 + h^2)^(-1/2) * (2ℓ)
dD/dw = (1/2)*(ℓ^2 + w^2 + h^2)^(-1/2) * (2w)
dD/dh = (1/2)*(ℓ^2 + w^2 + h^2)^(-1/2) * (2h)
Substituting the given values from the problem:
ℓ = 7 m
w = h = 2 m
Calculating:
(dD/dℓ) = (1/2)*(7^2 + 2^2 + 2^2)^(-1/2) * (2*7)
(dD/dw) = (1/2)*(7^2 + 2^2 + 2^2)^(-1/2) * (2*2)
(dD/dh) = (1/2)*(7^2 + 2^2 + 2^2)^(-1/2) * (2*2)
Simplifying:
(dD/dℓ) = (1/2)*(53)^(-1/2) * 14
(dD/dw) = (1/2)*(53)^(-1/2) * 4
(dD/dh) = (1/2)*(53)^(-1/2) * 4
Substituting the values of the rates of change:
dℓ/dt = 3 m/s
dw/dt = 3 m/s
dh/dt = -1 m/s
Calculating:
dD/dt = (1/2)*(53)^(-1/2) * 14 * 3 + (1/2)*(53)^(-1/2) * 4 * 3 + (1/2)*(53)^(-1/2) * 4 * (-1)
Simplifying:
dD/dt = (1/2)*(53)^(-1/2) * (14*3 + 4*3 - 4)
Calculating:
dD/dt ≈ 1.46
Therefore, at that instant, the rate at which the length of the diagonal of the box is changing is approximately 1.46 m/s.