Double-Checking Maths Homework for a 100% Semester Grade

[ertagon2]

So here is a thing. If I get at least 6/10 questions in this homework I will get 100% in this semester's homework which is equivalent to 20% of the maths module. So could you double check these questions? Also I wouldn't mind some help with q.3 I got a weird answer of 11,8 billion.
View attachment 7584
View attachment 7585

Workings:
https://imgur.com/a/wMUI3

 

[MarkFL]

2.) Correct.

3.) I would begin with:

\(\displaystyle \frac{1}{N}\d{N}{t}=\frac{1}{25}\) where \(\displaystyle N(0)=N_0\)

Integrate:

\(\displaystyle \int_{N_0}^{N(t)} \frac{du}{u}=\frac{1}{25}\int_0^t \,dv\)

\(\displaystyle \ln\left(\frac{N(t)}{N_0}\right)=\frac{t}{25}\)

\(\displaystyle N(t)=N_0e^{\Large{\frac{t}{25}}}\)

With $N_0=3,\,t=35$, we find:

\(\displaystyle N(35)=3e^{\Large{\frac{7}{5}}}\approx12.17\)

4.) I get a different answer. You've made a minor slip in your working, but your method is sound.

5.) Correct.

 

[ertagon2]

2.) Correct.

3.) I would begin with:

\(\displaystyle \frac{1}{N}\d{N}{t}=\frac{1}{25}\) where \(\displaystyle N(0)=N_0\)

Integrate:

\(\displaystyle \int_{N_0}^{N(t)} \frac{du}{u}=\frac{1}{25}\int_0^t \,dv\)

\(\displaystyle \ln\left(\frac{N(t)}{N_0}\right)=\frac{t}{25}\)

\(\displaystyle N(t)=N_0e^{\Large{\frac{t}{25}}}\)

With $N_0=3,\,t=35$, we find:

\(\displaystyle N(35)=3e^{\Large{\frac{7}{5}}}\approx12.17\)

4.) I get a different answer. You've made a minor slip in your working, but your method is sound.

5.) Correct.

Ah c=4 in q4
So f(2) = 2^2 + 2 + 4
= 10 ?
Thanks.

 

[MarkFL]

Ah c=4 in q4
So f(2) = 2^2 + 2 + 4
= 10 ?
Thanks.

Yes, that's what I got as well. :)

 

[ertagon2]

About q.3
This is what I did:

\(\displaystyle N(t)=Ae^{kt}\)

\(\displaystyle N(0)=Ae^{k(0)}=3\)

\(\displaystyle A=3\)

\(\displaystyle N(1)=3e^{k(1)}=3*1.04\)

\(\displaystyle 3e^{k}=3.12\)

\(\displaystyle e^{k}=\frac{3.12}{3}\)

\(\displaystyle (k)ln(e)=ln(1.04)\)

\(\displaystyle k=ln(1.04)\)

\(\displaystyle N(t)=3e^{ln(1.04)t}\)

\(\displaystyle N(35)=3e^{ln(1.04)(35)}\)

\(\displaystyle N(35)~11.84\)

Why is this wrong and what does\(\displaystyle \frac{1}{N}\frac{dN}{dt}\) mean in the context of the question?

 

[MarkFL]

About q.3
This is what I did:

\(\displaystyle N(t)=Ae^{kt}\)

\(\displaystyle N(0)=Ae^{k(0)}=3\)

\(\displaystyle A=3\)

\(\displaystyle N(1)=3e^{k(1)}=3*1.04\)

\(\displaystyle 3e^{k}=3.12\)

\(\displaystyle e^{k}=\frac{3.12}{3}\)

\(\displaystyle (k)ln(e)=ln(1.04)\)

\(\displaystyle k=ln(1.04)\)

\(\displaystyle N(t)=3e^{ln(1.04)t}\)

\(\displaystyle N(35)=3e^{ln(1.04)(35)}\)

\(\displaystyle N(35)~11.84\)

Why is this wrong and what does\(\displaystyle \frac{1}{N}\frac{dN}{dt}\) mean in the context of the question?

Your formula can be simplified to:

\(\displaystyle N(t)=N_0\left(1+\frac{1}{25}\right)^t\)

This is mathematically equivalent to an investment that is earning 4% annual interest compounded once per year. However, if the interest is continuously compounded then we have:

\(\displaystyle N(t)=N_0\lim_{n\to\infty}\left(\left(1+\frac{1}{25n}\right)^{nt}\right)=N_0e^{\large{\frac{t}{25}}}\)

Your model is discrete, whereas we are told to use a continuous growth model. :)