Exponential equation (simple?)

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Homework Help Overview

The discussion revolves around an exponential equation represented as x = Ae ^ kt, with specific initial conditions provided for x at t = 0 and t = 2. Participants are tasked with finding constants A and k, determining the value of x at t = 1.5, and calculating the time required for x to decay to 0.01. Additionally, a separate problem regarding the motion of a plane is introduced, involving a cubic function for distance as a function of time.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss substituting known values into the exponential equation to find A and k. There are attempts to clarify the process of finding initial conditions and derivatives related to the plane's motion. Questions arise about the application of differentiation and the implications of the results obtained.

Discussion Status

Some participants have offered guidance on how to approach the exponential equation and the plane's motion, while others express confusion regarding the differentiation process and its relevance to the problems at hand. Multiple interpretations of the plane's motion problem are being explored, particularly concerning the calculation of time when velocity reaches zero.

Contextual Notes

There is mention of a deadline for the plane problem, indicating time pressure on participants. Additionally, some participants question the appropriateness of tackling calculus-related problems without formal instruction in the subject.

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exponential equation (simple??)

x = Ae ^ kt

initally (at t = 0) x = 0.2 and when t = 2 then x = 1.5

a) Find A and k

b) Find x when t = 1.5

c) How long will it take x to decay to x = 0.01

Really struggling, any help would be greatly appreciated.
 
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1875 said:
x = Ae ^ kt

initally (at t = 0) x = 0.2 and when t = 2 then x = 1.5

a) Find A and k

b) Find x when t = 1.5

c) How long will it take x to decay to x = 0.01

Really struggling, any help would be greatly appreciated.

a) Plug in t=0 and x = 0.2 into the equation.
You get:
0.2 = A*e^0 or A=0.2
(this is like any exponential growth or decay function, at A = the initial value)
plug t = 2 into the equation and you get 0.2e^0.2k=1.5
ln|1.5/0.2|= 0.2k solve for k

b) plug t=1.5 in the equation above x(t) = 0.2e^kt (with whatever you solved for k)

c) same as b but put x = 0.01 in for x and solve for t
 
Last edited:


1875 said:
x = Ae ^ kt

initally (at t = 0) x = 0.2 and when t = 2 then x = 1.5

a) Find A and k

b) Find x when t = 1.5

c) How long will it take x to decay to x = 0.01

Really struggling, any help would be greatly appreciated.

Hi and welcome to PF.
Let's try a). When t=0 and x=0.2, can you write up the equation "x = Ae ^ kt"?
This should give you A.
 


Thank you, I completely understand it now, just would'nt click at first. I wonder while your online if you could help with;


at time t = 0 and position s = 0 a plane starts its descent into an airfield. From this point, the distance s in km as a function of time t in hours is given by;
s = 300 + 400t - 200t^3

determine
a) the inital velocity (km/hrs)
b) the acceleration after 1/2 hr
c) the time to when the velocity is zero and the distance traveled in that time

deadline tomorrow, pressure on
 


Am I right in thinking
a) 300 km/hr
b) 475 km/hr2
 


1875 said:
Thank you, I completely understand it now, just would'nt click at first. I wonder while your online if you could help with;


at time t = 0 and position s = 0 a plane starts its descent into an airfield. From this point, the distance s in km as a function of time t in hours is given by;
s = 300 + 400t - 200t^3

determine
a) the inital velocity (km/hrs)
b) the acceleration after 1/2 hr
c) the time to when the velocity is zero and the distance traveled in that time

deadline tomorrow, pressure on

First derivative is velocity so take that and plug in t=0
2nd derivative is acceleration, so take that and plug in t=0.5
 


Rebooter said:
First derivative is velocity so take that and plug in t=0
2nd derivative is acceleration, so take that and plug in t=0.5

I think I am OK on those parts its more the last question that I am having difficulty with;

the time to when the velocity is zero and the distance traveled in that time?
 


1875 said:
I think I am OK on those parts its more the last question that I am having difficulty with;

the time to when the velocity is zero and the distance traveled in that time?

Set the first derivative ds/dt = 0 = 400-600t^2 solve for t

Plug t into the first equation and solve for s
 


Rebooter said:
Set the first derivative ds/dt = 0 = 400-600t^2 solve for t

Plug t into the first equation and solve for s

for t I got

t = -400/600^2

Thanks again for your help, one last thing, I am just wondering how you came to this equation of 0 = 400 - 600t^2
 
  • #10


1875 said:
for t I got

t = -400/600^2

Thanks again for your help, one last thing, I am just wondering how you came to this equation of 0 = 400 - 600t^2

basic differentiation rules?
 
  • #11


Rebooter said:
basic differentiation rules?

I must apologise I am unaware of these rules and still struggling greatly on the last question.
 
  • #12


If you are not taking a Calculus course, why are you even attempting a problem involving finding the velocity from a position function? That requires Calculus.
 

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