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Math Help
#11
this is the full problem.


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#12
Would you need to multiply the top and bottom by (cos+1)? Hmm...

Edit: Nevermind. I don't know anything anymore lol.
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#13
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.
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#14
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?
Boston University: MS in Software Development (20/32 SH - 1 Course IP) | GPA: 3.88
Thomas Edison State University: BA in Liberal Studies GPA: 4.00 | AS in Natural Science and Mathematics (Computer Science, Mathematics) | Certificate in Electronics
Excelsior College: AAS in Technical Studies (Electronic/Instrumentation Technologies) - High Honors GPA: 3.79
Community College: AAS in Applied Science and Engineering Technology - Highest Honors GPA: 3.91 | AAS in Technical Studies (Computer Technology)

Epsilon Pi Tau International Honor Society Golden Key International Honor Society | Phi Theta Kappa Honor Society
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#15
(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!
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#16
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? 
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#17
(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.
Boston University: MS in Software Development (20/32 SH - 1 Course IP) | GPA: 3.88
Thomas Edison State University: BA in Liberal Studies GPA: 4.00 | AS in Natural Science and Mathematics (Computer Science, Mathematics) | Certificate in Electronics
Excelsior College: AAS in Technical Studies (Electronic/Instrumentation Technologies) - High Honors GPA: 3.79
Community College: AAS in Applied Science and Engineering Technology - Highest Honors GPA: 3.91 | AAS in Technical Studies (Computer Technology)

Epsilon Pi Tau International Honor Society Golden Key International Honor Society | Phi Theta Kappa Honor Society
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#18
(07-09-2019, 12:59 PM)MrBossmanJr Wrote:
(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.
But the correct answer to the question was √20 feet per second 
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#19
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.
Boston University: MS in Software Development (20/32 SH - 1 Course IP) | GPA: 3.88
Thomas Edison State University: BA in Liberal Studies GPA: 4.00 | AS in Natural Science and Mathematics (Computer Science, Mathematics) | Certificate in Electronics
Excelsior College: AAS in Technical Studies (Electronic/Instrumentation Technologies) - High Honors GPA: 3.79
Community College: AAS in Applied Science and Engineering Technology - Highest Honors GPA: 3.91 | AAS in Technical Studies (Computer Technology)

Epsilon Pi Tau International Honor Society Golden Key International Honor Society | Phi Theta Kappa Honor Society
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#20
(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?
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