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+---- Thread: Math Help (/Thread-Math-Help--32713)
RE: Math Help - Giantzebra - 06-25-2019
this is the full problem.
RE: Math Help - MrBossmanJr - 06-25-2019
Would you need to multiply the top and bottom by (cos+1)? Hmm...
Edit: Nevermind. I don't know anything anymore lol.
RE: Math Help - Giantzebra - 06-25-2019
I tried that .... I'm not sure .... Maybe there are some trig properties that I'm missing ...
I think you need to graph it but I'm not sure how.
RE: Math Help - MrBossmanJr - 06-25-2019
Is the answer -1/2?
I think you have to use L'Hospitals Rule.
f(0) = (cosx-1)/(xsinx)
so the denominator is basically 0*0 or x*x at this point which is x^2
cosx-1 can be written as -1+cosx or -(1-cosx)
Then you apply the rule and do the derivative of top and bottom
the top becomes -sinx and the bottom is 2x
the x's cancel and become 1's therefore being -1/2?
RE: Math Help - Giantzebra - 06-25-2019
(06-25-2019, 05:50 PM)MrBossmanJr Wrote: Is the answer -1/2?
I think you have to use L'Hospitals Rule.
f(0) = (cosx-1)/(xsinx)
so the denominator is basically 0*0 or x*x at this point which is x^2
cosx-1 can be written as -1+cosx or -(1-cosx)
Then you apply the rule and do the derivative of top and bottom
the top becomes -sinx and the bottom is 2x
the x's cancel and become 1's therefore being -1/2?
Ya - that seems like it makes sense. Thanks!
new math problem - Giantzebra - 07-09-2019
Objects moving in two dimensions can be hard to track, unless you use calculus to handle the motion. If an object is moving so its X coordinate follows the equation X = 2t + 1 and its Y coordinate follows the equation Y = t2 + 2, where t is the time in seconds, what is the object’s two-dimensional speed at t = 2 seconds?
How do you solve this problem?
RE: Math Help - MrBossmanJr - 07-09-2019
(07-09-2019, 11:59 AM)Giantzebra Wrote: Objects moving in two dimensions can be hard to track, unless you use calculus to handle the motion. If an object is moving so its X coordinate follows the equation X = 2t + 1 and its Y coordinate follows the equation Y = t2 + 2, where t is the time in seconds, what is the object’s two-dimensional speed at t = 2 seconds?
How do you solve this problem?
Don't you just differentiate both the equations? So x = t and y = 2t. Plug in t = 2 and you get (2,4).
It's asking for speed so you do differentiate. Speed (not a vector) is just the positive value of velocity (vector). The first equation should provide distance. Differentiating it once will provide velocity and another time will output acceleration.
RE: Math Help - Giantzebra - 07-09-2019
(07-09-2019, 12:59 PM)MrBossmanJr Wrote:But the correct answer to the question was √20 feet per second(07-09-2019, 11:59 AM)Giantzebra Wrote: Objects moving in two dimensions can be hard to track, unless you use calculus to handle the motion. If an object is moving so its X coordinate follows the equation X = 2t + 1 and its Y coordinate follows the equation Y = t2 + 2, where t is the time in seconds, what is the object’s two-dimensional speed at t = 2 seconds?
How do you solve this problem?
Don't you just differentiate both the equations? So x = t and y = 2t. Plug in t = 2 and you get (2,4).
It's asking for speed so you do differentiate. Speed (not a vector) is just the positive value of velocity (vector). The first equation should provide distance. Differentiating it once will provide velocity and another time will output acceleration.
RE: Math Help - MrBossmanJr - 07-09-2019
Ahhh, my bad forgot to mention. I found you the x and y values, but you need to do the pyrgahteka theorem to find the answer. 2^2 + 4^2 and square root the answer. Therefore, you come out with sq rt 20.
EDIT: My spelling is terrible and I can't remember the guy's name lol.
RE: Math Help - Giantzebra - 07-09-2019
(07-09-2019, 01:24 PM)MrBossmanJr Wrote: Ahhh, my bad forgot to mention. I found you the x and y values, but you need to do the pyrgahteka theorem to find the answer. 2^2 + 4^2 and square root the answer. Therefore, you come out with sq rt 20.
EDIT: My spelling is terrible and I can't remember the guy's name lol.
Thanks.
Do you also know how the extreme value theorem can be true if f(x) = x is continuous on every closed interval but has no minimum or maximum?
(07-09-2019, 01:29 PM)Giantzebra Wrote:(07-09-2019, 01:24 PM)MrBossmanJr Wrote: Ahhh, my bad forgot to mention. I found you the x and y values, but you need to do the pyrgahteka theorem to find the answer. 2^2 + 4^2 and square root the answer. Therefore, you come out with sq rt 20.
EDIT: My spelling is terrible and I can't remember the guy's name lol.
Thanks.
Do you also know how the extreme value theorem can be true if f(x) = x is continuous on every closed interval but has no minimum or maximum?
And, why do the Pythagorean theorem?