I need to do really well in Calculus . . .

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Im just starting the course, but is there anything in particular other than doing all the homework everynight that can help me achieve a very high mark? Ill probably come on this thread with questions i dont understand if that is alright?

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Type: Discussion • Score: 1 • Views: 1,250 • Replies: 17

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Sure it is.We are easy going on here.

In return you have to reciprocate if some of our answers are confusing.

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Ok, so my first question . . (i have just been on advanced functions for the last two months)

Say for the equation Y=x^2+4x-12 . . how would you find the equation of the tangent from first principles?

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Etruscia wrote:

Ok, so my first question . . (i have just been on advanced functions for the last two months)

Say for the equation Y=x^2+4x-12 . . how would you find the equation of the tangent from first principles?

First principles with our without calculus? In the former case, just differentiate.

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You cant use the x^2=2x rules or anything. You need to use the f(a+h)-f(a)/h equation

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Etruscia wrote:

You cant use the x^2=2x rules or anything. You need to use the f(a+h)-f(a)/h equation

Well, that looks like the basic pre-calculus way of finding a derivative as the limit of a difference quotient. Apply that to the original quadratic equation, and you will get a linear equation which is everywhere equal to the slope of the quadratic. Then you can use the point slope form of a straight line to find the tangent line to the quadratic at any point.

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ok, thanks. I thought anything with a derivitave made it calculus . .

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Etruscia wrote:

You cant use the x^2=2x rules or anything. You need to use the f(a+h)-f(a)/h equation

You've got the equation you need right here.

f(a+h) = (a+h)^2 + 4(a+h) -12 or a^2 + 2ah + h^2 + 4a + 4h -12
f(a) = a^2 + 4a - 12

The difference is: 2ah + h^2 + 4h

Divide by h and you get: 2a + h - 4

When you take the limit as h->0, you are just left with 2a + 4

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Isn't that the fundamental theorem of differentiation?

That is, the limit h->0 of [f(x+h)-f(x)]/h theorem?

Rap

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yup this is right ...........

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raprap wrote:

Isn't that the fundamental theorem of differentiation?

That is, the limit h->0 of [f(x+h)-f(x)]/h theorem?

Rap

Yes. I thought that is what the original question was. I remember having to do these when I first started calculus.

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This is only precalculus but . . .

How would you solve for variable "t" in this

Y=200(1.1)^t/3 Y=1000(.8)^t/4

When are they equal. Could you show me how to solve this algebraicly?

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To get t out of the exponent, you should take the logorithm of both sides. Remember that the log (a^b) = b*log(a) and log (cd) = log c + log d

log [200 (1.1)^(t/3)] = log 200 + (t/3) log(1.1)
log [1000 (.8)^(t/4)] = log 1000 + (t/4) log (.8)

Now you can combine these and solve with simple algebra.

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Oh man, thanks alot. I cant believe i didnt figure that out on the test lol! I still got 95% anyways;)

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IS there any tricks to solving related rate problems? SO far it is the hardest thing we have done, because one must find all the variables and create the equations ourselves. Any advice?

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ummm

related rates is the one where you make everything a derivative in respect to t right?

Sad thing is I did those like 3 weeks ago and I already forget what they are lol. They seemed pretty easy. Easier than optimization IMO. It helps to pay attention to the word problems and make sure you accurately identify every variable value given especially where dx/dt and dy/dt may be given as that will clue you in on how what equation you should solve.

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Heres a t question I found for you. (Not very difficult). Ironically it's about tea. (Drum Roll)

Yeah anyway, I'll post the answer in a few hours.

http://i22.photobucket.com/albums/b312/Red888/cupoftea.jpg

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Heres the solution to that question.

http://i22.photobucket.com/albums/b312/Red888/cupofteaA.jpg

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I need to do really well in Calculus . . .

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