Calculating dQ/dt: Chain Rule Applied

Click For Summary
SUMMARY

The discussion focuses on calculating the derivative dQ/dt using the chain rule for the function Q = √(x² + y)e^t, where x = √(1 - e^(-2t)) and y = 2 - e^(-2t). The initial calculations yield dQ/dt = √(3 - 2e^(-2t))e^t + (4e^(-t))/(2√(3 - 2e^(-2t))). Participants suggest that while the substitution of x and y is valid, it may be more beneficial to keep the variables in the expression until the end to potentially simplify the final result.

PREREQUISITES
  • Understanding of the chain rule in calculus
  • Familiarity with exponential functions and their derivatives
  • Knowledge of square root functions and simplification techniques
  • Ability to manipulate algebraic expressions involving multiple variables
NEXT STEPS
  • Review the application of the chain rule in multivariable calculus
  • Learn techniques for simplifying expressions involving square roots
  • Explore the implications of variable substitution in calculus problems
  • Study the properties of exponential functions and their derivatives
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus, as well as educators looking for examples of applying the chain rule in derivative calculations.

jamesbob
Messages
63
Reaction score
0
Question: Let
Q = \sqrt{x^2 + y}e^t​
where (for t > or = 0)
x = \sqrt{1 - e^{-2t}}​
and
y = 2 - e^{-2t}​

Using the chain rule calculate dQ/dt, expressing your answer in as simple a form as possible. My work so far

Subbing in values of x and y:

Q = \sqrt{1 - e^{-2t} + 2 - e^{-2t}}e^t = \sqrt{3 - 2e^{-2t}}e^t

Now applying the product and chain rule:

u = (3 - 2e^{-2t})^{\frac{1}{2}} \left chain \left rule \Rightarrow u = z^{\frac{1}{2}} \left z = 3 - 2e^-2t

\frac{du}{dz} = \frac{1}{2\sqrt{z}}, \left \frac{dz}{dt} = 4e^-2t

So \frac{du}{dt} = \frac{4e^{-2t}}{2\sqrt{3 - 2e^{-2t}}}

As for v, v = e^t, \left \frac{dv}{dt} = e^t

So \frac{dQ}{dt} = \sqrt{3 - 2e^{-2t}}e^t + \frac{4e^{-t}}{2\sqrt{3 - 2e^{-2t}}}

Is this right so far and can it be simplified further?
 
Physics news on Phys.org
jamesbob said:
Question: Let
Q = \sqrt{x^2 + y}e^t​
where (for t > or = 0)
x = \sqrt{1 - e^{-2t}}​
and
y = 2 - e^{-2t}​

Using the chain rule calculate dQ/dt, expressing your answer in as simple a form as possible.


My work so far

Subbing in values of x and y:

Q = \sqrt{1 - e^{-2t} + 2 - e^{-2t}}e^t = \sqrt{3 - 2e^{-2t}}e^t

Now applying the product and chain rule:

u = (3 - 2e^{-2t})^{\frac{1}{2}} \left chain \left rule \Rightarrow u = z^{\frac{1}{2}} \left z = 3 - 2e^-2t

\frac{du}{dz} = \frac{1}{2\sqrt{z}}, \left \frac{dz}{dt} = 4e^-2t

So \frac{du}{dt} = \frac{4e^{-2t}}{2\sqrt{3 - 2e^{-2t}}}

As for v, v = e^t, \left \frac{dv}{dt} = e^t

So \frac{dQ}{dt} = \sqrt{3 - 2e^{-2t}}e^t + \frac{4e^{-t}}{2\sqrt{3 - 2e^{-2t}}}

Is this right so far and can it be simplified further?

Looks good to me. Theoretically you would want to remove the square root from the denominator and add the two expressions, but I personally don't see that that actually simplifies anything.

-Dan
 
Thanks very much :smile:
 
Subbing in values of x and y:
While a perfectly valid technique, I think part of the point of this problem is not to do such a substitution.
 
Hmm, i see, so would i just do the chian rule keeping x and y in, and then sub values in right at the end?
 
Or maybe not even sub in the values at all -- the final result might (or might not) look nicer in terms of x and y!
 

Similar threads

Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
5
Views
2K
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
5
Views
2K