Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
dmcampos
Group Title
18.01SC Single Variable Calculus:
Problem Set 1.
Question 1F7c:
The result I got was
(3mgr) / ( (1+r^2)^(7/4) )
but the solutions say it is
(3mgr) / ( (1+r^2)^(5/2) )
Can anyone explain to me what I did wrong?
I used both the chain rule and the quotient rule, and after doing it, I got this:
[  mg * ( (3/2) * (1 + r^2)^(1/2) * 2r ) ] / ( (1 + r^2)^(9/4) )
 5 months ago
 5 months ago
dmcampos Group Title
18.01SC Single Variable Calculus: Problem Set 1. Question 1F7c: The result I got was (3mgr) / ( (1+r^2)^(7/4) ) but the solutions say it is (3mgr) / ( (1+r^2)^(5/2) ) Can anyone explain to me what I did wrong? I used both the chain rule and the quotient rule, and after doing it, I got this: [  mg * ( (3/2) * (1 + r^2)^(1/2) * 2r ) ] / ( (1 + r^2)^(9/4) )
 5 months ago
 5 months ago

This Question is Closed

phi Group TitleBest ResponseYou've already chosen the best response.1
the power rule might be the easiest way: \[ d \ x^n = n x ^{n1} dx \] \[ \frac{d}{dr} \frac{m g}{\left( 1+r^2\right)^{\frac{3}{2}} } = \frac{d}{dr} m g \left( 1+r^2\right)^{\frac{3}{2}} \] m and g are constants, so we can move them outside the derivative \[ m g \frac{d}{dr}\left( 1+r^2\right)^{\frac{3}{2}} \\ =  \frac{3}{2}m g\left( 1+r^2\right)^{\frac{3}{2} 1} \frac{d}{dr}\left( 1+r^2\right)\\ =  \frac{3}{2}m g\left( 1+r^2\right)^{\frac{5}{2}} 2 r\\ = \frac{3mgr}{\left( 1+r^2\right)^{\frac{5}{2}}} \]
 5 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
You can use the quotient rule (though normally when the numerator is a constant you would not bother) your answer is close: \[ \frac{mg \cdot \frac{3}{2} (1+r^2)^{\frac{1}{2}} \cdot 2 r}{(1 + r^2)^{\frac{9}{4}}} \] The "bug" is that the square of the denominator is \[ \left( (1+r^2)^\frac{3}{2} \right)^2= (1+r^2)^{\frac{3}{2}\cdot 2}= (1+r^2)^3\] if we use the correct denominator we have \[ \frac{mg \cdot \frac{3}{\cancel{2}} (1+r^2)^{\frac{1}{2}} \cdot \cancel{2} r}{(1 + r^2)^{3}} \\ = \frac{3mg r}{(1 + r^2)^{3\frac{1}{2}}} \\ = \frac{3mg r}{(1 + r^2)^\frac{5}{2}} \] which is the correct result
 5 months ago

dmcampos Group TitleBest ResponseYou've already chosen the best response.0
Thanks a lot, your answer was perfect, and I figured out my mistake. (and I'm sorry I didn't give many details, I was just going update the post and you had already answered) It was really simple: I assumed \[((1+r^{2})^{\frac{ 3 }{ 2 }})^{2} = (1+r^{2}) ^{\frac{ 9 }{ 4 }}\]instead of \[(1+r^{2})^{3}\]which is obviously the correct way.
 5 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.