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Dodo1
Dear, Math-talent! The slope of the tangent line to the graph of the function y=3x^3 at the point (381) is limx=>3 (3x^3−81)/(x-3.) By trying values of x near 3, find the slope of the tangent line
Take you calculator and try x=2.9, 2.99, 2.999, to get a good idea where this is going to...
ok,Thank you. I have a calculator with me.. but why x=2.9??
oh i get it beacuse its near 3....
Well, you want to know the limit for x --> 3. You are not supposed to do all kinds of math trickery to do this, but just try a few values near 3. So 2.9, or 3.1 and then even closer: 2.99 or 3.01, or any other value, as long as it is "near" 3...
Once I gt all answers from equation. what shall I do?
Can you guess what the number is that you are getting closer and closer to?
hmmmm I will try put 2.9 , 2.999 in the equation first Please hold!
Ok, So I got 78.33 for x=2.9 and 80.7303 for x=2.99... I guess it gets closer to 81?
cuze at the point (3, 81)
It does, so it's a safe bet 81 is the slope of the tangent line!
(3,81) begin the point on the graph is in itself not of importance. By coincidence, the y-coordinate is also 81.
Omg, thats prety cool! thank you very much :) are you faimilar with Tagent Velocity:
You mean the velocity of a moving point along a curve is the tangent vector? Or is it a book? (havent't got a clue here..)
I could be familiar with the stuff in the book ;)
cool. The displacement (in meters) of a particle moving in a straight line is given by s=3t^3 where t is measured in seconds. 1)Find the average velocity of the particle over the time interval [7,10]. 2)Find the (instantaneous) velocity of the particle when t=7
Do you know the difference between average velocity and instantaneous velocity?
mmm No. I have no idea. av velo =displacement/ time elapsed?
It is, so that one is easy. Just calculate (s(10-s(7))/(10-7).
For instanteneous velocity, you'll have to do the limit-stuff as in the first problem. Can you write down the limit?
so s(10)-s(7)/3? for the fist one?
Yes, because s(10-s(7) is the distance you've travelled.
and jusy plug s=3t^3 in s?
Don't get what you're saying there...
s(10)=3*10³=3000, s(7)=3*7³=1029 (3000-1029)/3=657 m/s (you may not be using meters, but feet...)
For instanteneous velocity, I have to do limit thing? x=>7?
Write down the limit:\[v(7)=\lim_{t \rightarrow 7}\frac{ s(t)-s(7) }{ t-7 }\] and try numbers near 7.
OK, I tired to 6.99. s(6,99)=(6.99)^3*3=1024.59 S(7)= 1029 1024.59-1029/1029-7?
You have to divide by t-7, which is 6.99-7=-0.01
441 is good. Just use brackets...
Oh, now I get it ;) If you set t=6.999 you'll be even closer to 441, so that will be the answer
Oh! So limit is how you get closer to x or function?
Yes, it is the value of a formula as you get closer and closer to a certain number. Only, because of the special kind of formula, it is not possible to try that number directly. As in the formula above: if you want the speed at t=7, you can't just put t=7 and get the right answer. If you try it, you'll get 0/0, which is undefined.
It is possible however, to do more advanced calculations with the formula to get the limit. It is called differentiation... Once you know how to do that, you don't need to try all these cumbersome numbers "near" to a certain value!
Wow, thank you so much for your detailed explaination. It is really fun doing math with you!
Thanks for the compliment! I really appreciate it.