Find the equation of the form y = Ce^kt if it passes through two specific points

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Homework Help Overview

The problem involves finding the equation of the form y = Cekt that passes through the points (0, 4) and (5, 1/2). The context is centered around exponential functions and their properties.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to differentiate the equation y = Cekt and relate it to the linear equation y = mx + b, but expresses confusion about how to incorporate the second point into their reasoning. They also question the appropriate use of derivatives versus integrals in this context.

Discussion Status

Some participants question the original poster's interpretation of the problem and clarify that the goal is to find the constants C and k by substituting the given points into the equation. There is acknowledgment of a misunderstanding, and a participant confirms the correct approach to substituting the coordinates.

Contextual Notes

The original poster seems to be struggling with the application of derivatives and integrals in relation to the problem, indicating a potential gap in understanding the relationship between the exponential function and its parameters.

lude1
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Homework Statement


Find the equation of the form y = Cekt if it passes through (0, 4) and (5, 1/2)

Homework Equations


y = mx+b?


The Attempt at a Solution


When I see "find the equation", my instant reaction is to find the derivative of y = Cekt. Once I have the derivative (or the slope of the equation I'm finding) I can plug it into y = mx+b. Plug in one set of points I was given to find "b", and plug "b" back into the original equation along with my derivative/slope. However, I don't know what to do with the second pair of points. Do I use it to check my work? Nevertheless, if my reasoning is correct, I'm not sure how to find the derivative of y = Cekt because there are three unknown variables. This is what I have so far:

y = C(ekt)' + ektC'
y = C(ekt)(kt)' + 0

I stopped here because I noticed it didn't look right. If I continue with this derivative, I end up with 0 because the derivative of kt will be 0 (assuming they are both variables) making the first part of the equation zero, and like the first part, the derivative of C is zero (assuming it is a variable) making the second part of the equation zero (thus having 0 + 0).

Which leads to another question: when do I find the derivative and when do I find the integral? In my other homework problems, I was given the slope and a pair of points. However, this problem was solved by integrating the slope and getting another equation. Then, with the integral, the book plugged in the point and got the particular solution they asked for.
 
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I don't understand what you are attempting to do. They gave you y = Cekt and are asking for the equation if it passes through (0, 4) and (5, 1/2).

They want you to find C and k.
 
Oh.. I must have misinterpreted the problem. So that means I plug in my x coordinates for t and y coordinates for y, right?
 
lude1 said:
Oh.. I must have misinterpreted the problem. So that means I plug in my x coordinates for t and y coordinates for y, right?

Yes.
 
Mmk. Thank you for clarifying the problem for me!
 

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