Derivatives Problem (Calculus I)

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SUMMARY

The discussion focuses on finding the derivative of the area of a rectangle, represented by the formula A=bh, with respect to time (t). Participants clarify that both the base (b) and height (h) are functions of time, necessitating the use of the product rule and chain rule for differentiation. The correct approach to derive dA/dt involves recognizing that dA/dt = b(dh/dt) + h(db/dt), where dh/dt and db/dt represent the rates of change of height and base over time, respectively. Misunderstandings regarding the presence of the variable t and the application of derivative rules are addressed.

PREREQUISITES
  • Understanding of basic calculus concepts, specifically derivatives.
  • Familiarity with the product rule for differentiation.
  • Knowledge of the chain rule for derivatives.
  • Ability to interpret functions of multiple variables.
NEXT STEPS
  • Study the application of the product rule in calculus.
  • Learn about the chain rule and its implications in multi-variable functions.
  • Practice problems involving derivatives of area for different geometric shapes.
  • Explore real-world applications of derivatives in physics and engineering contexts.
USEFUL FOR

Students studying calculus, particularly those tackling derivatives in relation to geometric shapes, as well as educators seeking to clarify derivative concepts in a classroom setting.

PotentialE
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Homework Statement


find d/dt for a rectangle.

Homework Equations


A=bh
product rule for derivatives (the first times the derivative of the second plus the second times the derivative of the first)
Chain rule for derivatives

The Attempt at a Solution


If b is a constant, then I know that dA/dh = b (this was the previous problem in which I could solve)
my issue with the current problem is that this variable t that I'm supposed to take the derivative with respect to is not in the problem- so how would I go about doing this?

My attempt:
d/dt (A=bh)
dA/dt = bh
dA/dt = (b(dA/dh)) + (h(dA/db))
My only problem with this solution is that I think there should be a d(b or h)/dt on the right side of the equation because of the chain rule

could someone please explain where/why the variable t comes in and what the correct answer is? Thank you!
 
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PotentialE said:
find d/dt for a rectangle.

I have no clue what this means. Can you quote the full problem as it appears in your book?
 
that's my problem too... that IS the quote!

"Find d/dt for a rectangle" and then the formula given is A=bh
so find d/dt for A=bh
 
Well, that makes no sense. I advice asking your teacher for more explanations.
 
On a previous problem:
Find d/dt for a square

and the answer was:
d/dt (A=s2)

dA/dt = 2s (ds/dt)

perhaps you understand this one? (I don't)
well i know the derivative of s2 = 2s. so we've taken the derivative of A with respect to t... so perhaps multiplying 2s by ds/dt is part of the chain rule or something... I don't really know where ds/dt came from
 
PotentialE said:

Homework Statement


find d/dt for a rectangle.


Homework Equations


A=bh
product rule for derivatives (the first times the derivative of the second plus the second times the derivative of the first)
Chain rule for derivatives

The Attempt at a Solution


If b is a constant, then I know that dA/dh = b (this was the previous problem in which I could solve)
my issue with the current problem is that this variable t that I'm supposed to take the derivative with respect to is not in the problem- so how would I go about doing this?

My attempt:
d/dt (A=bh)
dA/dt = bh
dA/dt = (b(dA/dh)) + (h(dA/db))
My only problem with this solution is that I think there should be a d(b or h)/dt on the right side of the equation because of the chain rule

could someone please explain where/why the variable t comes in and what the correct answer is? Thank you!
Assume that b and h are both functions of time.

There should be no derivatives of A on the right hand side of the following:

dA/dt = (b(dA/dh)) + (h(dA/db))

Use the product rule to find \displaystyle \frac{d}{dt}(b\,h)\ .[/itex
 

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