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http://www.mei.org.uk/files/papers/c406ja_ssi4c0.pdf

Can anyone help me with 8ii please. a is 2.5 which I have worked out but in order to work out k ,why cant I just sub in (2.5-1.6) instead of dx/dt and 2.5 instead of a and x in order to find k. It gives me an answer of -0.36 which is wrong

I was told to just sub into the main equation rather than the rate of change one so I did that with some success until I ran into the following question

http://www.mei.org.uk/files/papers/c409jn_gegd.pdf 7ciii

I get how to find A but to find K you have to sub into the rate of change equation this time. So confused, I can do all the integration parts just having problems with which equation to sub into.

Whats even more confusing is for the q7iii, they use the rate of change after 1hr which is -1.5 but use the initial value of 98. surely 96.5 should be used instead

Thanks in advance

Can anyone help me with 8ii please. a is 2.5 which I have worked out but in order to work out k ,why cant I just sub in (2.5-1.6) instead of dx/dt and 2.5 instead of a and x in order to find k. It gives me an answer of -0.36 which is wrong

I was told to just sub into the main equation rather than the rate of change one so I did that with some success until I ran into the following question

http://www.mei.org.uk/files/papers/c409jn_gegd.pdf 7ciii

I get how to find A but to find K you have to sub into the rate of change equation this time. So confused, I can do all the integration parts just having problems with which equation to sub into.

Whats even more confusing is for the q7iii, they use the rate of change after 1hr which is -1.5 but use the initial value of 98. surely 96.5 should be used instead

Thanks in advance

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(Original post by

http://www.mei.org.uk/files/papers/c406ja_ssi4c0.pdf

Can anyone help me with 8ii please. a is 2.5 which I have worked out but in order to work out k ,why cant I just sub in (2.5-1.6) instead of dx/dt and 2.5 instead of a and x in order to find k. It gives me an answer of -0.36 which is wrong

**helpme456**)http://www.mei.org.uk/files/papers/c406ja_ssi4c0.pdf

Can anyone help me with 8ii please. a is 2.5 which I have worked out but in order to work out k ,why cant I just sub in (2.5-1.6) instead of dx/dt and 2.5 instead of a and x in order to find k. It gives me an answer of -0.36 which is wrong

Whereas, (2.5-1.6)/(1-0) gives the average rate of change over the first year.

You also made another mistake in reasoning, when you put x=2.5 for when t=1. When t=1, x=1.6.

All you need to do is use the first equation they give you (which it seems you already used to find a). Let x=1.6, and t=1.

I was told to just sub into the main equation rather than the rate of change one so I did that with some success until I ran into the following question

http://www.mei.org.uk/files/papers/c409jn_gegd.pdf 7ciii

I get how to find A but to find K you have to sub into the rate of change equation this time. So confused, I can do all the integration parts just having problems with which equation to sub into.

Whats even more confusing is for the q7iii, they use the rate of change after 1hr which is -1.5 but use the initial value of 98. surely 96.5 should be used instead

Thanks in advance

http://www.mei.org.uk/files/papers/c409jn_gegd.pdf 7ciii

I get how to find A but to find K you have to sub into the rate of change equation this time. So confused, I can do all the integration parts just having problems with which equation to sub into.

Whats even more confusing is for the q7iii, they use the rate of change after 1hr which is -1.5 but use the initial value of 98. surely 96.5 should be used instead

Thanks in advance

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(Original post by

I'm guessing you thought dx/dt = (2.5-1.6)/(1-0)? This doesn't work because dx/dt is the instantaneous rate of change (of x with respect to t); it tells you the rate of change at a particular point in time (you can think of it graphically as the gradient of the tangent to that point).

Whereas, (2.5-1.6)/(1-0) gives the average rate of change over the first year.

You also made another mistake in reasoning, when you put x=2.5 for when t=1. When t=1, x=1.6.

All you need to do is use the first equation they give you (which it seems you already used to find a). Let x=1.6, and t=1.

If you understood what I typed above then I think you should be able to do this. If not, I'll reply after some sleep

**scherzi**)I'm guessing you thought dx/dt = (2.5-1.6)/(1-0)? This doesn't work because dx/dt is the instantaneous rate of change (of x with respect to t); it tells you the rate of change at a particular point in time (you can think of it graphically as the gradient of the tangent to that point).

Whereas, (2.5-1.6)/(1-0) gives the average rate of change over the first year.

You also made another mistake in reasoning, when you put x=2.5 for when t=1. When t=1, x=1.6.

All you need to do is use the first equation they give you (which it seems you already used to find a). Let x=1.6, and t=1.

If you understood what I typed above then I think you should be able to do this. If not, I'll reply after some sleep

what about the next question http://www.mei.org.uk/files/papers/c409jn_gegd.pdf 7ciii

is 1.5 the instantaneous change after 1hr?

also would it be 3 after 2hrs or is there no way of finding out

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#4

(Original post by

I understand what you are saying, I just worked out the average change, not the instantaneous so I should have just subbed into the main equation.

what about the next question http://www.mei.org.uk/files/papers/c409jn_gegd.pdf 7ciii

is 1.5 the instantaneous change after 1hr?

**helpme456**)I understand what you are saying, I just worked out the average change, not the instantaneous so I should have just subbed into the main equation.

what about the next question http://www.mei.org.uk/files/papers/c409jn_gegd.pdf 7ciii

is 1.5 the instantaneous change after 1hr?

No, why did you think that? There's nothing in the question that says 1 hour.

-1.5 is the initial rate, as they tell you. So at the start when t=0, dθ/dt=-1.5. How did you find A, if you didn't do this? (Also, remember the rate is negative because the temperature is going down.)

also would it be 3 after 2hrs or is there no way of finding out

How would you begin to work out the rate after 2 hours?

Spoiler:

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The obvious thing to try would be the equation for the rate, which is dθ/dt = -k(θ-θ

_{0})Now what do you need to proceed?

Spoiler:

Show

You know k and θ

_{0}are constants (which you know the values of), which only leaves θ. θ is the temperature, and it's a variable which is dependent on time. So, there's a particular θ after 2 hours - can you find this?
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(Original post by

-1.5 is the initial rate, as they tell you. So at the start when t=0, dθ/dt=-1.5. How did you find A, if you didn't do this? (Also, remember the rate is negative because the temperature is going down.)

No, it wouldn't be 3 - again, why did you think this? I think you might be working with these concepts without knowing/considering what they actually mean (in my experience this is the case with most students), so let's try to clarify.

How would you begin to work out the rate after 2 hours?

Now what do you need to proceed?

**scherzi**)**No, why did you think that? There's nothing in the question that says 1 hour.**-1.5 is the initial rate, as they tell you. So at the start when t=0, dθ/dt=-1.5. How did you find A, if you didn't do this? (Also, remember the rate is negative because the temperature is going down.)

No, it wouldn't be 3 - again, why did you think this? I think you might be working with these concepts without knowing/considering what they actually mean (in my experience this is the case with most students), so let's try to clarify.

How would you begin to work out the rate after 2 hours?

Spoiler:

Show

The obvious thing to try would be the equation for the rate, which is dθ/dt = -k(θ-θ

_{0})Now what do you need to proceed?

Spoiler:

Show

You know k and θ

_{0}are constants (which you know the values of), which only leaves θ. θ is the temperature, and it's a variable which is dependent on time. So, there's a particular θ after 2 hours - can you find this?
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#6

(Original post by

It says The initial rate of temperature loss is 1.5◦F per hour.

**helpme456**)It says The initial rate of temperature loss is 1.5◦F per hour.

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