Implicit Function: Box Dimensions & Rates of Change

In summary: If you take the derivative you get dL/dt= 1/2(ℓ^2+ w^2+ h^2)^{-1/2}(2ℓ dℓ/dt+ 2w dw/dt+ 2h dh/dt). Now, at time t, ℓ= 4, w= 9, and h= 9 so L= (4^2+ 9^2+ 9^2)^{1/2}= (16+ 81+ 81)^{1/2}= 178^{1/2}= 13.34. So we have dL/dt= 1/2(13.34)^{-
  • #1
Physicsnoob90
51
0

Homework Statement


The length ℓ, width w, and height h of a box change with time. At a certain instant the dimensions are ℓ = 4 m and
w = h = 9 m, and ℓ and w are increasing at a rate of 1 m/s while his decreasing at a rate of 6 m/s. At that instant find the rates at which the following quantities are changing.

(A) The Volume
(B) The Surface Area
(C) the length of a diagonal (round two decimals places)

Homework Equations


The Chain rule, Partial Derivative

The Attempt at a Solution


I already found A (ans: -99 m^3/s) and B (ans: -94 m^2/s)
for C:
1. i differentiated the formula L^2= ℓ^2+w^2+h^2 to 2L(dL/dt) = 2ℓ(dℓ/dt) + 2w(dw/dt) + 2h (dh/dt)
2. let dℓ/dt = dw/dt = 1 m/s and dh/dt = -6 m/s
3. my answer came as 2L(dL/dt) = 82 but I'm completely lost after this part
 
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  • #2
Physicsnoob90 said:

Homework Statement


The length ℓ, width w, and height h of a box change with time. At a certain instant the dimensions are ℓ = 4 m and
w = h = 9 m, and ℓ and w are increasing at a rate of 1 m/s while his decreasing at a rate of 6 m/s. At that instant find the rates at which the following quantities are changing.

(A) The Volume
(B) The Surface Area
(C) the length of a diagonal (round two decimals places)

Homework Equations


The Chain rule, Partial Derivative

The Attempt at a Solution


I already found A (ans: -99 m^3/s) and B (ans: -94 m^2/s)
for C:
1. i differentiated the formula L^2= ℓ^2+w^2+h^2 to 2L(dL/dt) = 2ℓ(dℓ/dt) + 2w(dw/dt) + 2h (dh/dt)
2. let dℓ/dt = dw/dt = 1 m/s and dh/dt = -6 m/s
3. my answer came as 2L(dL/dt) = 82 but I'm completely lost after this part
At time t, you know what the values of w, h, and l are. You want to solve 2L (dL/dt) = 82 for dL/dt.
 
  • #3
SteamKing said:
At time t, you know what the values of w, h, and l are. You want to solve 2L (dL/dt) = 82 for dL/dt.
would i be able to find L by square rooting (w,h,ℓ) and then multiplying it with the 2?

update: i manage to figure out the equation by doing just that. Thanks for your help!
 
Last edited:
  • #4
You could but if you implicit differentiation you shouldn't! The length of the diagonal is given by [itex]L= (ℓ^2+ w^2+ h^2)^{1/2}[/itex].
 
Last edited by a moderator:

Related to Implicit Function: Box Dimensions & Rates of Change

1. What is an implicit function?

An implicit function is a mathematical relationship between two or more variables, where one variable is not explicitly expressed in terms of the other variable(s). This means that the relationship between the variables is only implied, rather than being stated directly.

2. How are box dimensions and rates of change related in an implicit function?

In an implicit function, the dimensions of a box (such as height, length, and width) are related to each other through a rate of change. This means that as one dimension of the box changes, the other dimensions will also change in a predictable way.

3. How do you find the rate of change in an implicit function?

The rate of change in an implicit function can be found using the concept of differentiation. This involves taking the derivative of the implicit function with respect to the variable of interest. The resulting derivative will give the rate of change at a specific point on the graph of the implicit function.

4. What is the significance of understanding implicit functions in the field of science?

Implicit functions are used in many scientific fields, such as physics, engineering, and economics. Understanding implicit functions allows scientists to model and analyze complex systems, predict how one variable will change in relation to another, and make informed decisions based on this information.

5. Can implicit functions be used to solve real-world problems?

Yes, implicit functions can be used to solve a variety of real-world problems. For example, they can be used to determine optimal dimensions for a box to minimize material costs, or to predict the growth rate of a population based on various factors. Implicit functions are a powerful tool in problem-solving and decision-making in many scientific disciplines.

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