# Rates associated calculus - derivative

• masterchiefo
In summary: The symbol ##s## is used in the SI system of units (mks) for length, and as a lowercase ##s## for time.You're doing it again - writing things that aren't true.dS/dh ≠ 6h2 and 6h2 ≠ 12hdS/dt = 6cm2/hour6/ (12h) = 1/(2h)I finally get it! You had to make me work it out myself. Alright, now I can do this actually. Thanks
masterchiefo

## Homework Statement

Growth is observed for a cubic crystal. Initially the height of the cube is 1 cm . the surface of the cube increases at a rate of 6 cm2 / hour.

Question: Calculate dh/dt

## Homework Equations

ds/dt = ds/dh * dh/dt

## The Attempt at a Solution

ds/dt = 6 cm2 / hour
ds/dh = h*h = 2*h

I do not understand how to get to that answer thank you.

masterchiefo said:

## Homework Statement

Growth is observed for a cubic crystal. Initially the height of the cube is 1 cm . the surface of the cube increases at a rate of 6 cm2 / hour.

Question: Calculate dh/dt

## Homework Equations

ds/dt = ds/dh * dh/dt

## The Attempt at a Solution

ds/dt = 6 cm2 / hour
ds/dh = h*h = 2*h
You're trying to do too much all at once here. h * h ≠ 2h

What is the equation of S in terms of h?
What is the rate of change of S with respect to h?
masterchiefo said:
If you want this to be interpreted as what you intended (but not as what you wrote), you need parentheses: 6/(2h).
masterchiefo said:
The answer I should get: 1/2h
That answer is not correct, at least with regard to what you wrote, which is the same as (1/2)h. If you want to indicate that 2h is in the denominator, put parentheses around it, like this: 1/(2h)
Or write it using LaTeX, like so ##\frac {1} {2h}##
What I wrote is this: # #\frac {1} {2h}# #, without the extra spaces.
masterchiefo said:
I do not understand how to get to that answer thank you.
See above.

Mark44 said:
You're trying to do too much all at once here. h * h ≠ 2h

What is the equation of S in terms of h?
What is the rate of change of S with respect to h?
If you want this to be interpreted as what you intended (but not as what you wrote), you need parentheses: 6/(2h).
That answer is not correct, at least with regard to what you wrote, which is the same as (1/2)h. If you want to indicate that 2h is in the denominator, put parentheses around it, like this: 1/(2h)
Or write it using LaTeX, like so ##\frac {1} {2h}##
What I wrote is this: # #\frac {1} {2h}# #, without the extra spaces.
See above.
Surface in terms of h = height * height = h2
d/dh = 2*h

Not really sure I understand how different I should do this...the answer is 1/(2h).

masterchiefo said:
Surface in terms of h = height * height = h2
d/dh = 2*h
Not really sure I understand how different I should do this...
Earlier you wrote "ds/dh = h*h = 2*h"
This is incorrect, since 1) dS/dh ≠ h*h and 2) h*h ≠ 2h. It appeared that you were munging together two concepts into one mush that didn't make much sense.
My aim was to get you to write two equations: one for S in terms of h, and the other for ##\frac{dS}{dt}##

In this thread you wrote "d/dh = 2*h", which is incorrect for a technical reason. d/dh is not the derivative - it is the differentiation operator - something that you apply to a function to get a derivative. What you probably meant was dS/dh = 2h. (Note that for surface area, S is usually used. Lower case s is usually used for distance, as in s = (1/2)gt2. I have no idea what s stands for, but it's used a lot for this purpose in many physics books.)
In your relevant equation, you have dS/dt = dS/dh * dh/dt
Use this equation and the quantities you know to solve for dh/dt.

Mark44 said:
Earlier you wrote "ds/dh = h*h = 2*h"
This is incorrect, since 1) dS/dh ≠ h*h and 2) h*h ≠ 2h. It appeared that you were munging together two concepts into one mush that didn't make much sense.
My aim was to get you to write two equations: one for S in terms of h, and the other for ##\frac{dS}{dt}##

In this thread you wrote "d/dh = 2*h", which is incorrect for a technical reason. d/dh is not the derivative - it is the differentiation operator - something that you apply to a function to get a derivative. What you probably meant was dS/dh = 2h. (Note that for surface area, S is usually used. Lower case s is usually used for distance, as in s = (1/2)gt2. I have no idea what s stands for, but it's used a lot for this purpose in many physics books.)
In your relevant equation, you have dS/dt = dS/dh * dh/dt
Use this equation and the quantities you know to solve for dh/dt.
dS/dh = 6*h2 =12h
dS/dt = 6cm2/hour

6/ (12h) = 1/(2h)

masterchiefo said:
dS/dh = 6*h2 =12h
You're doing it again - writing things that aren't true.
dS/dh ≠ 6h2 and 6h2 ≠ 12h
masterchiefo said:
dS/dt = 6cm2/hour

6/ (12h) = 1/(2h)

Mark44 said:
(Note that for surface area, S is usually used. Lower case s is usually used for distance, as in s = (1/2)gt2. I have no idea what s stands for, but it's used a lot for this purpose in many physics books.)
I always saw the use of ##s## as a math book thing. :) According to Wikipedia, the reason ##s## is used is because the Latin word for length is spatium.

## 1. What is the concept of rates associated with calculus and derivatives?

The concept of rates associated with calculus and derivatives refers to the relationship between the change in one quantity and the change in another quantity. In calculus, this is represented by the derivative, which measures the instantaneous rate of change of a function at a specific point. It allows us to understand how quickly a function is changing and in what direction.

## 2. How is the derivative calculated in rates associated calculus?

The derivative is calculated by finding the slope of the tangent line to a function at a given point. This is done by using the limit definition of the derivative, which involves taking the limit as the change in the independent variable approaches zero.

## 3. What is the significance of derivatives in real-world applications?

Derivatives have a wide range of applications in the real world. They are used in physics to describe motion and forces, in economics to model supply and demand, in engineering to optimize designs, and in many other fields. Derivatives allow us to understand how different quantities are related and how they change over time or in response to certain inputs.

## 4. Can derivatives be negative?

Yes, derivatives can be negative. A negative derivative indicates that the function is decreasing at a certain point, while a positive derivative indicates that the function is increasing. The magnitude of the derivative also tells us the rate of change, with a larger magnitude indicating a faster rate of change.

## 5. How do higher-order derivatives relate to rates associated calculus?

Higher-order derivatives, such as the second or third derivative, represent the rate of change of the derivative itself. They can give us information about the curvature of a function and how it is changing. In some cases, higher-order derivatives may be used to model more complex real-world phenomena that cannot be accurately described by a first-order derivative.

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