Rates associated calculus - derivative

[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

My answer: 6/2hThe answer I should get: 1/2h

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

 

[Mark44]

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?

My answer: 6/2h

If you want this to be interpreted as what you intended (but not as what you wrote), you need parentheses: 6/(2h).

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.

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

See above.

 

[masterchiefo]

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).

 

[Mark44]

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.

 

[masterchiefo]

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)

 

[Mark44]

dS/dh = 6*h² =12h

You're doing it again - writing things that aren't true.
dS/dh ≠ 6h² and 6h² ≠ 12h

dS/dt = 6cm²/hour

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

 

[vela]

(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.