aivantettet26
  • aivantettet26
Inverse Trigonometric Functions : Differentiation Problem
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
aivantettet26
  • aivantettet26
1 Attachment
anonymous
  • anonymous
|dw:1378459457916:dw|
aivantettet26
  • aivantettet26
1 Attachment

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
This is my imagining of it.
anonymous
  • anonymous
|dw:1378459530270:dw|
anonymous
  • anonymous
Notice how \[ \tan\theta = \frac x {50} \]
aivantettet26
  • aivantettet26
yes, and then \[\arctan \frac{ x }{ 50 } = \]
aivantettet26
  • aivantettet26
\[\arctan \frac{ x }{ 50 } =\theta\]
anonymous
  • anonymous
That is the answer to the first part, but you don't want to use it to solve the second part.
aivantettet26
  • aivantettet26
uhmm.. okay okay
anonymous
  • anonymous
For the second part, you want do differentiate: \[ \tan (\theta)=\frac x {50} \]with respect to \(x\)
anonymous
  • anonymous
\(\color{blue}{\text{Originally Posted by}}\) @wio For the second part, you want do differentiate: \[ \tan (\theta)=\frac x {50} \]with respect to \(x\) \(\color{blue}{\text{End of Quote}}\) I mean with respect to \(t\).
anonymous
  • anonymous
Can you do implicit differentiation?
aivantettet26
  • aivantettet26
inversely or trigo method?
aivantettet26
  • aivantettet26
yes
anonymous
  • anonymous
Okay so differentiate it implicitly.
aivantettet26
  • aivantettet26
\[\sec ^2(\theta)( d \theta) = \frac{ dx/dt }{ 50 }\]
aivantettet26
  • aivantettet26
\[(1+\tan^2(\theta))(d \theta)=\frac{ dx/dt }{ 50 }\]
aivantettet26
  • aivantettet26
am I on the right track?
anonymous
  • anonymous
You went a bit too far
anonymous
  • anonymous
\[ \sec^2(\theta)\frac{d\theta}{dt} = \left(\frac{1}{50}\right) \frac {dx}{dt} \]
anonymous
  • anonymous
This is good enough.
anonymous
  • anonymous
Now we need to solve for \(dx/dt\).
aivantettet26
  • aivantettet26
\[50\sec^2(45)\frac{ d \theta }{ dt } = \frac{ dx }{ dt }\]
anonymous
  • anonymous
What is \[ \frac{d\theta}{dt} \]
anonymous
  • anonymous
Hint: 30 revolutions per minute
aivantettet26
  • aivantettet26
yes
anonymous
  • anonymous
How fast is that in radians?
aivantettet26
  • aivantettet26
\[60\pi?\]
anonymous
  • anonymous
How many radians per second?
aivantettet26
  • aivantettet26
188.5?
anonymous
  • anonymous
It's just \(\pi\)
anonymous
  • anonymous
So \[ \frac{d\theta}{dt} = \pi \]
aivantettet26
  • aivantettet26
pi / 30?
aivantettet26
  • aivantettet26
whoa?
anonymous
  • anonymous
Okay I'm wrong never mind. It can remain in rad per minute. Thus \[ \frac{d\theta}{dt} = 60\pi \]
aivantettet26
  • aivantettet26
\[50\sec2(45)(60\pi)=dx/dt\]
anonymous
  • anonymous
Yeah. Now just simplify.
anonymous
  • anonymous
The answer is in feet per minute
aivantettet26
  • aivantettet26
\[6000\pi?\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.