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Stuck on a Maths Question

10) A substance is decaying in such a way that its mass, m kg, at time t years from now is given by the formula m=240e^-0.04t.
a) Find the time taken for the substance to halve its mass (i've done this one)
b) Find the value of t for which the mass is decreasing at a rate of 2.1kg per year. (i'm stuck on this one).

This is for exponential models and it's the new Math spec. Bear in mind that I have not done differentiation so I have to work out the answer using log, ln, e, etc.
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Dat1Guy
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Original post by tyronemarsh
10) A substance is decaying in such a way that its mass, m kg, at time t years from now is given by the formula m=240e^-0.04t.
a) Find the time taken for the substance to halve its mass (i've done this one)
b) Find the value of t for which the mass is decreasing at a rate of 2.1kg per year. (i'm stuck on this one).

This is for exponential models and it's the new Math spec. Bear in mind that I have not done differentiation so I have to work out the answer using log, ln, e, etc.


You need to differentiate the function and make it equal to 2.1 using ln won't allow you to calculate this.

Differentiating y=e^2x = 2e^2x
Y=2e^4x^2 = 16xe^4x^2 for example
(edited 6 years ago)
Original post by tyronemarsh
10) A substance is decaying in such a way that its mass, m kg, at time t years from now is given by the formula m=240e^-0.04t.
a) Find the time taken for the substance to halve its mass (i've done this one)
b) Find the value of t for which the mass is decreasing at a rate of 2.1kg per year. (i'm stuck on this one).

This is for exponential models and it's the new Math spec. Bear in mind that I have not done differentiation so I have to work out the answer using log, ln, e, etc.


That just screams differentiation to me.

Otherwise, you can approximate it by sketching this graph and observing where the gradient is (roughly) -2
There is a way of solving for t without using differentiation. I figured it out last night!

Given that any rate of change is A=kAe^-k

We can solve for t in this way:
Mass is changing at a rate of -2.1 (decreasing)

-0.04x240e^-0.04t=-2.1
-9.6e^-0.04t=-2.1
e^-0.04t=-2.1/-9.6
-0.04tlne=-ln(2.1/9.6)
0.04t=ln(2.1/9.6)
t=37.99... which is the correct answer :smile:

Just thought i'd tell you guys haha
Original post by tyronemarsh
There is a way of solving for t without using differentiation. I figured it out last night!

Given that any rate of change is A=kAe^-k

We can solve for t in this way:
Mass is changing at a rate of -2.1 (decreasing)

-0.04x240e^-0.04t=-2.1
-9.6e^-0.04t=-2.1
e^-0.04t=-2.1/-9.6
-0.04tlne=-ln(2.1/9.6)
0.04t=ln(2.1/9.6)
t=37.99... which is the correct answer :smile:

Just thought i'd tell you guys haha


That uses differentiation though... a derivative with respect to tt (time) IS a rate of change over time, since ddt(240e−0.04t)=−0.04⋅240e−0.04t\frac{d}{dt}(240e^{-0.04t})=-0.04\cdot 240e^{-0.04t} as you have. So really, you haven't answered your own original post with this, unless you were given the derivative but you didn't include it in the original post.
(edited 6 years ago)
Original post by RDKGames
That uses differentiation though... a derivative with respect to tt (time) IS a rate of change over time, since ddt(240e−0.04t)=−0.04⋅240e−0.04t\frac{d}{dt}(240e^{-0.04t})=-0.04\cdot 240e^{-0.04t} as you have. So really, you haven't answered your own original post with this, unless you were given the derivative but you didn't include it in the original post.


I'm guessing my teacher taught me how to differentiate it indirectly haha. As it was to the power of -0.04t I just used that constant and put it to multiply 240e lol. The textbook explains such rules like this without differentiation

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