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Car depreciation problem

  • Thread starter jai6638
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URGENT Basic Calc Help needed

Q3) If a car depreciates continuously at a rate of 23.5%, when ( years ) will a 25,000 dollar car be worth $12,000 ?

A= pe^rt
(12,000)/(25000)= e^(.235t)
Ln ( 12,000/25000)=.235t
T= .0038

This answer does not seem right. Where am I going wrong?

Q2) Find the vertical and horizontal Asymptotes of f(x) = (3x) / ( 2x-4)

Vertical Asymptote = 2
Horizontal Asymptote= dunno.. I cant use limits here directly since the exponents of X are the same. If i tried dividing them so as to get a proper form, it doesnt seem to work out.



Q3)

Find the equation of the normal line at x=2 of the function f(x) = x^2-5x+4

X=2
Y=-2 ( plugged in 2 into the function )

Y=mx+b
-2=2m+b

How would I find the slope and B? Is the slope -1?


Q4) Find the average rate of change between 2 and -1 of

f(x) = x^2+ 3x

I found the derivative to be 2x+3 and plugged in 2 and -1 into the derivative to get -2 and 10.

10=11/2 +b = 9/2 = b

Is this the answer? y= -11/4x + 9/2
 
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Answers and Replies

TD
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1) I think this should work: [tex]25000\left( {1 - 0.235} \right)^x = 12000[/tex]

2) Your VA is correct but why can't you use limits for the HA? If the exponent in the nominator is one more than the one of the denominator, you may have any line as asymptote but HA appear when the highest exponents are equal.
Try taking the limit to + and - infinity.

3) If you can find the tangent, you're almost finished, since you just have to take the perpendicular line then. (with m' = -1/m)

4) Don't you need to find a numerical answer? I'd say the avarage change between x = a and x = b would be (f(b)-f(a))/(b-a)
 
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2) Your VA is correct but why can't you use limits for the HA? If the exponent in the nominator is one more than the one of the denominator, you may have any line as asymptote but HA appear when the highest exponents are equal.
Try taking the limit to + and - infinity.
2) well the text book said that I cant use limits for HA when the exponent of X in num and den are equal. Anyhow, I tried to do so, and the X cancel out thus giving me a limit of 3/2 .. is that correct?

1) It is bein compounded continuously so how can I use the standard thre equation that you listed. However, I tried it anyways and got an answer of 2.73 years... dunno if this is correct.
 
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For Question 3: is y=2x-5 the answer? ( i plugged in 2 into the derivative of the equation and then put it in point soope form )
 
TD
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jai6638 said:
2) well the text book said that I cant use limits for HA when the exponent of X in num and den are equal. Anyhow, I tried to do so, and the X cancel out thus giving me a limit of 3/2 .. is that correct?

1) It is bein compounded continuously so how can I use the standard thre equation that you listed. However, I tried it anyways and got an answer of 2.73 years... dunno if this is correct.

For Question 3: is y=2x-5 the answer? ( i plugged in 2 into the derivative of the equation and then put it in point soope form )
2) Yes, you have a HA (both at the right as left) at y = 3/2

1) I wasn't 100% sure about that one, someone else may verify.

3) That doesn't seem correct. Your slope at (2,f(2)) is -1 so the perpendicular direction is 1. So the line should be y - f(2) = 1*(x-2)
 
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y - f(2) = 1*(x-2)
that cnat be the final equation ?

Is the final equation y= 2x since Y seems to equal 2 according to the equoted equation and if you plug it in point slope form ( 2-2= b ) you would get y=2x.
 
TD
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y = f(2) would be -2. So the line becomes: y = x - 4
 
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hmm I see.. so you dont have to find a derivitive eh?

thanks .. need some help with the other answers though
 
TD
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I did use the derivative to find the slope (of the tangent line).
If that slope is m, the perpendicular slope of the normal line is m' = -1/m.
 
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wont the perp slope be (-1/3) then?

Damn im confused. I'd really appreciate it ifyou could please write down the steps for what you did for q3..
 
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jai6638 said:
Damn im confused. I'd really appreciate it ifyou could please write down the steps for what you did for q3..
Would you mind if I tried?

The Bob (2004 ©)
 
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sure.............
 
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[tex]f(x) = x^2 - 5x + 4[/tex]
[tex]f(2) = 2^2 - (5 \times 10) + 4 = 4 - 10 + 4 = -2[/tex]
We now have the place where the normal line will cross, at (2, -2).

Differentiate the function to get:

[tex]f'(x) = 2x - 5[/tex]

Put in the value of x, which is 2, and we get the gradient at that point on the curve:

[tex]f'(2) = (2 \times 2) - 5 = -1[/tex]

To find the perdendicular gradient to this, we have to find the missing value (as we know that f'(2) is equal to m1):

[tex]m_1 \times m_2 = -1[/tex]

[tex]-1 \times m_2 = -1[/tex]

[tex]m_2 = \frac{-1}{-1} = 1[/tex]

We now have everything we need for the second equation of a straight line:

[tex]y - y_1 = m (x - x_1)[/tex]

We know that y1 is -2 and that x1 is 2. We now know that m (the gradient) is 1 (for the normal line to that equation at x = 2).

Plug it all in and you get:

[tex]y - y_1 = m (x - x_1)[/tex]

[tex]y - (-2) = 1 (x - 2)[/tex]

[tex]y + 2 = x - 2[/tex]

[tex]\Rightarrow y = x - 4[/tex]

I hope that all makes sense. Do say if not.

The Bob (2004 ©)
 
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I'm relatively new to differenciation so i'd like to know why are we finding the perpendicular gradient since we want to find the value at a particular point?

Also, how is (m1)(m2)=-1?

THanks much. really appreciate it guys..
 
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jai6638 said:
I'm relatively new to differenciation so i'd like to know why are we finding the perpendicular gradient since we want to find the value at a particular point?

Also, how is (m1)(m2)=-1?

THanks much. really appreciate it guys..
Sorry for the late reply. :blushing:

The differentiated line is the tangent to the line. The normal line, which you want, is perpendicular to the tangent of that point.

The way to work out the perdendicular gradient is (m1)(m2) = -1. Take the line y = x. The gradient is going to be 1. The line perpendicular to this one is y = -x (through the origin); but why? That equation. To the m's multiply to -1, the other m value (as one of them is 1) must be -1. The same is for your posted question. It is then used in the Second Equation for Straight Lines.

Did any of that make any sense at all?

The Bob (2004 ©)
 
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Yeah I did... essentially, its coz the slope of a perp line is the negative inverse ( -1/m ) eh?

Cool.. thanks much.. Could you also confirm the other answers please?
 
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jai6638 said:
Yeah I did... essentially, its coz the slope of a perp line is the negative inverse ( -1/m ) eh?

Cool.. thanks much.. Could you also confirm the other answers please?
Ok, but bear in mind that I am very tired and that I am going to bed after this post.

1. TD knocked that one on the head in post 2
2. [tex]x = 2[/tex] and [tex]y = \frac{3}{2}[/tex]
3. y = x - 4
4. I believe it to be 4 but that might be wrong.

Now I must sleep. It is 13 minutes to 1 in the morning here. :smile:

The Bob (2004 ©)

P.S. Hope that helps.
 
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thanks much

Can anyone confirm the answer for question 4?
 
TD
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jai6638 said:
thanks much

Can anyone confirm the answer for question 4?
Using (f(b)-(a))/(b-a) as I said, I agree with Bob that is should be 4.

I had to log off last night but I see Bob helped you further, hope it's all clear now :smile:
 
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I tried puttig it in that form and i got -2+10 / 2+1 = 8/3

doh.. need to go for my exam now.... thanks anyways.
 
TD
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jai6638 said:
I tried puttig it in that form and i got -2+10 / 2+1 = 8/3

doh.. need to go for my exam now.... thanks anyways.
f(x) = x^2+ 3x between 2 and -1

f(-1) = -2
f(2) = 10

=> (f(2)-f(-1))/(2-(-1)) = (12)/(3) = 4

Good luck with your exam!
 
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Thanks much... damn im trying to call my school office and no one picking up.. ! I hope they arent taking a day off or something!
 
TD
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It would give you more time to prepare, not your fault ;)

If you have any more questions, don't hesitate to ask for help!
 
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finally got through... gonna take it at 1 pm... thank god!


thanks.
 
TD
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Good luck then!
 

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