- #1
Your original post makes no sense. If you want any help you'll have to organise it into something meaningful.cemtu said:dN0/dt = 2 is given thus N0 is 2t. There is no problem there sir.
I put the whole question and its solution.PeroK said:Your original post makes no sense. If you want any help you'll have to organise it into something meaningful.
I know how to derive however our professor just replaced N with dN/dt (LHS) and N0 with dN0/dt (RHS). I don't know how. He didnt took the derivative as the way you and I want.etotheipi said:Are you sure you don't mean ##A_{0} = \left[-\frac{dN}{dt} \right]_{t=0} = 2##?
Of course, start off with the equation you wrote of ##N = N_{0}e^{-\lambda t}##. Then you can derive both sides with respect to ##t## to obtain a relation pertaining to the activities.
How would you derive the RHS wrt ##t##? Or perhaps more specifically, what can we do about the constant out the front?
cemtu said:I know how to derive however our professor just replaced N with dN/dt (LHS) and N0 with dN0/dt (RHS). I don't know how. He didnt took the derivative as the way you and I want.
Hereetotheipi said:I can't speak for what your professor was trying to imply however I can say that ##\frac{dN_{0}}{dt}## is zero, as is the case for all constants. It's essentially a meaningless statement.
What do you get if you derive both sides of ##N = N_0 e^{-\lambda t}## with respect to ##t##?
I suspect this might be an initial value problem. But, you've misunderstood what's been given as initial values.cemtu said:Here
cemtu said:Here
Sir, thank you.etotheipi said:You've done the differentiation right, ##\frac{dN}{dt} = -\lambda N_{0} e^{-\lambda t} \implies A = A_{0}e^{-\lambda t}## since ##A = \lambda N##.
You must stop using ##\frac{dN_{0}}{dt}##. It is wrong. I strongly suspect the intended meaning is the rate of change of ##N## at ##t=0##, however we might just denote this (negative) ##A_{0}## (or that more clunky expression in post #7).
You should be able to just plug in the ratio of the activities.
cemtu said:Sir, thank you.
This...PeroK said:Without seeing the problem stated clearly, we are all just guessing.
A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It is essentially the slope of a tangent line to the function at that point.
Derivatives are used in many areas of science and engineering to model and analyze various processes and phenomena. They are particularly useful in understanding how variables change over time and in optimizing functions.
The most common method for calculating a derivative is using the power rule, which states that the derivative of a function is equal to the exponent multiplied by the coefficient, with the exponent decreased by 1. There are also other rules and techniques, such as the chain rule and product rule, for calculating derivatives of more complex functions.
Derivatives have many applications in science and engineering, such as in physics (to calculate velocity and acceleration), economics (to determine marginal cost and revenue), and biology (to model population growth). They are also used in machine learning and artificial intelligence to optimize algorithms and make predictions.
Some common mistakes when working with derivatives include forgetting to apply the chain rule or product rule, making algebraic errors, and not simplifying the final answer. It is important to carefully follow the rules and double-check calculations to avoid these mistakes.