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Dear, Mathtalent!
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)/(x3.) By trying values of x near 3, find the slope of the tangent line
 one year ago
 one year ago
Dear, Mathtalent! 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)/(x3.) By trying values of x near 3, find the slope of the tangent line
 one year ago
 one year ago

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ZeHanzBest ResponseYou've already chosen the best response.1
Take you calculator and try x=2.9, 2.99, 2.999, to get a good idea where this is going to...
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
ok,Thank you. I have a calculator with me.. but why x=2.9??
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
oh i get it beacuse its near 3....
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
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...
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
Once I gt all answers from equation. what shall I do?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
Can you guess what the number is that you are getting closer and closer to?
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
hmmmm I will try put 2.9 , 2.999 in the equation first Please hold!
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
Ok, So I got 78.33 for x=2.9 and 80.7303 for x=2.99... I guess it gets closer to 81?
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
cuze at the point (3, 81)
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
It does, so it's a safe bet 81 is the slope of the tangent line!
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
(3,81) begin the point on the graph is in itself not of importance. By coincidence, the ycoordinate is also 81.
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
Omg, thats prety cool! thank you very much :) are you faimilar with Tagent Velocity:
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
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..)
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
I could be familiar with the stuff in the book ;)
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
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
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
Do you know the difference between average velocity and instantaneous velocity?
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
mmm No. I have no idea. av velo =displacement/ time elapsed?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
It is, so that one is easy. Just calculate (s(10s(7))/(107).
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
For instanteneous velocity, you'll have to do the limitstuff as in the first problem. Can you write down the limit?
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
so s(10)s(7)/3? for the fist one?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
Yes, because s(10s(7) is the distance you've travelled.
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
and jusy plug s=3t^3 in s?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
Don't get what you're saying there...
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
s(10)=3*10³=3000, s(7)=3*7³=1029 (30001029)/3=657 m/s (you may not be using meters, but feet...)
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
For instanteneous velocity, I have to do limit thing? x=>7?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
Write down the limit:\[v(7)=\lim_{t \rightarrow 7}\frac{ s(t)s(7) }{ t7 }\] and try numbers near 7.
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
OK, I tired to 6.99. s(6,99)=(6.99)^3*3=1024.59 S(7)= 1029 1024.591029/10297?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
You have to divide by t7, which is 6.997=0.01
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
441 is good. Just use brackets...
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
Oh, now I get it ;) If you set t=6.999 you'll be even closer to 441, so that will be the answer
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
Oh! So limit is how you get closer to x or function?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
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.
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
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!
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
Wow, thank you so much for your detailed explaination. It is really fun doing math with you!
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
Thanks for the compliment! I really appreciate it.
 one year ago
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