# Help for my daughter

1. Mar 5, 2008

### Tompman

Hi,

(e^0.03t)((50e^0.07t-10)/k)^-0.5 dt

Yours sincerely,

Tom.

2. Mar 5, 2008

### Gib Z

Doing that problem analytically is probably not possible, but if we very vaguely approximate the integrand, we can see what it is for most practical purposes.

Your integral is simplifies to $$\sqrt{\frac{k}{50}} \int^{\infty}_0 \frac{e^{0.03t}}{e^{0.07t} - 0.2} dt$$.

The next step takes away a lot of accuracy, but its the best i can really think of that doesnt take forever to do. We pretend we don't see the -0.2 in the denominator lol.

If we do that and evaluate the integral, we should get $$5\sqrt{ \frac{k}{2} }$$.

3. Mar 5, 2008

### HallsofIvy

What class is your daughter in that she would have a problem like that?

4. Mar 5, 2008

### Tompman

Gib Z,
Thank you. I have sent this to my daughter.

HallsofIvy,
My daughter is doing economics in university and this has something to do with the reserves of oil in the world.

5. Mar 5, 2008

### nicksauce

Using maple the integral

$$\int^{\infty}_0 \frac{exp(0.03t)}{exp(0.07t) - 0.2}dt$$

is approximately 27.07839303

6. Mar 6, 2008

### Gib Z

O thats pretty good then =] When we ignored that -0.2, it was 25.

7. Mar 6, 2008

### TheoMcCloskey

Am I missing something, or shoudn't the integral portion be

$$\int^{\infty}_0 \frac{e^{0.03t}}{\sqrt{e^{0.07t} - 10}}dt$$

8. Mar 6, 2008

### Tompman

Yes Theo, having spoken to my daughter I think you are correct. What would the answer be then? All replies are very welcome.

9. Mar 6, 2008

### nicksauce

According to maple, the integral

$$\int_0^{\infty}\frac{exp(0.03t)}{\sqrt{50exp(0.07t)-10}}$$

is approximately 28.48936147

10. Mar 7, 2008

### Gib Z

Sorry my bad >.<" Maybe thats why I was so surprised that my 'approximation' was quite close, because I only ignored 0.2 instead of a 10. If we try to ignoring thing again I'm sure it would be quite a bit off.

11. Mar 7, 2008

### TheoMcCloskey

I'm still looking at this, but I'm not done. However, I don't think the value is that high. Initially, I'm getting something roughly half this value.

more to come.

12. Mar 7, 2008

### TheoMcCloskey

After further review, I indeed get results that agree with nicksauce, ie, The integral is approx equal to 28.489361...