# A few questions

1. May 24, 2006

### danago

Hi. I have exams coming up soon, and i was given a practice exam to...practice with. Anyway, there are a few questions i am unsure about, and would appreciate it if anyone could please tell me if i have done them correctly.

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1. The first term of a geometric series is 5, and the common ratio is 3. Find:
a) The seventh term
b) The sum of the first 9 terms
c) The number who's value is 885735
d) The smallest value for which $$T_n>7000000$$

For part a, it asks for the seventh term. Since in the question it states that it is a series, does that mean the 5th term is the sum of the first 5 terms of the progression, or is the 5th term of the progression? so would it be:

$$T_n=ar^{n-1}$$
$$=5\times 3^{6}$$
$$=3645$$

or

$$S_n=\frac{a(1-r^n)}{1-r}$$
$$=\frac{5(1-3^7)}{1-3}$$
$$=5465$$

which one?

Part b i think i did correctly. I got 49205 as the sum of the first 9 terms.

Part c, again, the same problem. Does it want a term number which is actuall 885735, or the term number where all of the previous terms add up to 885735?

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2. A patch of lillies in a pond begin to grow in a circular shape. In the first year, they form a diameter of 4 meters.

Each following year, the diameter increased by 75% of the previous years increase. Therefore, in the second year, an increase of 3m occured, and in the next, 2.25m and so on.

a) How much did the lilly pad diamater grow in the 5th year?
b) What was the diameter after the 15th year?
c) If the width of the river was 17 meters, would the lilly patch ever touch the sides of the river? why not?

I think i did this correctly, but would like some confirmation.

a) $$T_5=4\times 0.75^{4}=1.266m$$
b) $$S_15=frac{4(0.75^15-1)}{0.75-1}=15.786m$$
c) No, the lilly patch will never touch the sides of the river. Since the common ratio is in between 1 and -1, the series converges to a single number, but never actually reaches it (since term number never reaches infinity), in this case, 16, which is less than 17.

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3. Differenciate the following:
a)$$y=5x^3+x^2-8x+26$$
b)$$y=(3x^2+5)^-2$$
c)$$y=(\sqrt{2x^2+3})^3$$
d)$$y=\frac{2}{2x-3}$$

a) $$\frac{dy}{dx}=15x^2+2x-8$$

b) For the second, i used the chain rule, and said:
$${\rm y = u}^{{\rm - 2}}$$
$${\rm u = 3x}^{\rm 2} + 5$$

$$\displaylines{ \frac{{dy}}{{dx}} = nu^{n - 1} u' \cr = - 12x(3x^2 + 5)^{ - 3} \cr}$$

c) I used the chain rule again, and got
$$\frac{{dy}}{{dx}} = 9x\sqrt {2x^2 + 3}$$

d) For d, i used the quotient rule. My answer was:
$$\frac{{dy}}{{dx}} = \frac{{ - 4}}{{(2x - 3)^2 }}$$

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4. Given that $$f(x)=3x^2-a$$, and f'(2)=12, is it possible to calculate the value of a? Why / why not?

If it was given that f(1)=5, calculate the value of a.

I said no, it was not possible to calculate a. Since if the equation was differentiated in respect to a, for all values of x in f'(x), -1 would result, so f(x) must have been differentiated with respect to x, which means that a will have no effect at all on f'(x) (since it is considered a constant), thus making it impossible to find a value for a.

The next part asks me to calculate a, giving me more information. I just substituted x=1 into the function, and made it equal 5, then solved for a, which gave me -2.

Have i explained myself properly, and calculated for a correctly?

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5. If a stone is dropped from a bridge, the distance it falls in meters is given by the equation:
$$s=4.88t^2$$

where t is time in seconds since it was dropped.

a) how far will it fall in 1.5 seconds?
b) what will be the velocity after 3 seconds?
c) how many seconds will it take to reach a velocity of 60m/s?

I simply substituted x=1.5 into the equation and got an answer of 10.98m.

For part b, it differentiated to get:
$$\frac{ds}{dt}=9.76t$$

Substituted 3 into that to get a velocity of 29.28m/s after 3 seconds. For the third part, part c, i made the derivative of the distance equal to 60, and solved for t, giving me 6.15 seconds.

Thats all for now. Even if i have done them correctly, any information on more efficient ways, or extra information i should include would be great. Thanks very much.
Dan.

Last edited: May 24, 2006
2. May 24, 2006

### Hootenanny

Staff Emeritus
Hi there danago, I'll do my best to answer your questions but I haven't got that much time at the moment, any I don't get round to I'll try to come back later and finnish them off. I would like to say that the presentation of this thread is excellent.

Question One

(a)The question states that a = 1 and r = 3; and asks for the fifth term, it does not say the "sum of the first five terms, therefore it just want the value of the fifth term.

(c) I would say the same as the above, it wants the term who's value is 885735.

Question Two

Looks good to me.

Question Three

3.(a) Correct
3.(b) Correct
3.(c) You may want to check this one. HINT: $y=(\sqrt{2x^2+3})^3 = (2x^2 + 3)^\frac{3}{2}$
3.(d) Correct

Question Four

This looks good except you said that f'(x) is constant, but it is not! f'(2) is constant; but f'(x) = 6x and is therefore variable (remember that x2 does not have a constant gradient?). If the gradient was constant, then the function would be linear not quadratic. The second part is correct.

Question Five

Spot on.

~H

Last edited: May 24, 2006
3. May 24, 2006

### danago

Thanks very much for the reply.

With question 3c, is the answer $$6x\sqrt {2x^2 + 3}$$? I did a stupid mistake when i was doing it, and instead of writing the derivative of 'u' as 4x, i did it as 6x.

And with question 4, what i was saying is that if it was differentiated in respect to 'a', that f'(x) would be constant. Wouldnt $$\frac{{dy}}{{da}} = - 1$$?

4. May 25, 2006

### Hootenanny

Staff Emeritus
Yep, you got it.

Sorry, that would be a misinterpreation on my part. I thought you were stating the f'(x) is constant. Yeah, it makes more sense now when I re read it.

Once again I would like to congratualte you on the presentation and effort that you have put into your thread. I have given it a 5/5 rating in a trial rating system the Homework Helpers and Science Advisors are running.

~H

5. May 25, 2006

### danago

Thanks for the rating, even though im not too sure what it means

And thanks very much for your time. I greatly appreciate it

6. May 25, 2006

### Hootenanny

Staff Emeritus