Use a calculator to get one answer then draw cos(x) and think about how to work out the second
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The Math Help Thread
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Use a calculator to get one answer then draw cos(x) and think about how to work out the second
Lion Heart
Banned
I add pie + 0.75 and I get 3.89. Is that right and the book is off by .03?
No
You're trying to solve the equation cosx = -0.75, which has two solutions between 0 and 2pi. Plugging in x = arccos(-0.75) will give you the solution which is between 0 and pi. (2.42...) The other solution will be the same distance away from pi as the first solution (since the graph of the cosine function is symmetric around the line x = pi), so you can find it by subtracting the first solution from pi and adding the difference back to pi.
-0.75 is the y coordinate on the plot and you are looking for the two values for x
Lion Heart
Banned
Ah I think I understand. This is exclusive to cos I imagine, which is what screwed me up. Thanks guys.
opticalmace
Member
Is it a Calc I course (i.e. differential calculus)? What is your math background?
Lion Heart
Banned
I'm given a sinusoidal equation, y=0.9sin(pi/2)x where x is time and y is meters.
It asks for the time at 0.5 meters above ground, how do I do this? The answers are 0.37s and 1.63s.
I input 0.5 for y but after I solve for x I get .556 so something is wrong on my end.
It asks for the time at 0.5 meters above ground, how do I do this? The answers are 0.37s and 1.63s.
I input 0.5 for y but after I solve for x I get .556 so something is wrong on my end.
Have an Elliptic Curves exam tomorrow and haven't been given proper past papers. I actually understand the concepts pretty well but have no idea what to expect on the exam. Hoping I can blag it with lots of drawings of parallelograms and lattices.
I think you still have pi/2 hanging around
I think you still have pi/2 hanging around
you sure that x is not inside the sin bracket? not really a sin equations when sin(pi/2) is just a numerical value.
Northeastmonk
Gold Member
http://www.wolframalpha.com/
I used that website for Calculus I and Physics 1 & 2. I was lost with Calculus because my prep for it was pre-Calc and it was mainly trig and geometry. To some this might of helped, but I was still lost.
It takes some time, but Calculus I is not that hard. Learn the concepts forwards and backwards. You have to memorize a few formulas like the multiplication, division, and subtraction methods. Then learning how to solve derivatives.
They sort of teach Calculus in order. I think the last thing we did were integrals and the fundamental theorem of calculus.
Watch youtube videos and visit Khan Academy. Its very easy because like I said Calculus, to me, was taught in a specific order. Know what it means when X approaches 0 or infinity and etc.
Limits - http://www.youtube.com/watch?v=lCSk4Df2x-A
^ huge help if you understand that concept.
I did very well in Physics 1 (2 I did good, but its nothing like Physics 1 bc Physics 2 is all electricity) because I learned a lot from my Calculus class.
http://www.youtube.com/watch?v=GGQngIp0YGI
See if there is an online syllabus for the course. Grab the textbook and review the chapters.
Lion Heart
Banned
Wolfram Alpha says:
http://www.wolframalpha.com/input/?i=0.5=0.9sin(pi/2x)
essentially, sin(pi x /2) = 0.555... = 5/9. Just take the sine inverse and find both solutions.
So you got 0.5/0.9 = sin(xpi/2), right? I'm gonna use 5/9 because I'm lazy.
The solutions to the above are:
xpi/2 = 2npi + pi - arcsin(5/9)
xpi/2 = 2npi + arcsin(5/9),
with n being an integer. Apply n = 0, because you want the first time your equation hits the first and the second quadrants and you're set.
TyphoonStruggle
Member
Physics question:
A back-packer grabs her 75 N pack and drags it with constant speed at an angle of 30 degrees to the horizontal. The force she applies to the pack is 40 N. What is the magnitude of the kinetic frictional force that is acting on the pack?
A. 20 N
B. 28 N
C. 35 N
D. 40 N
My solution:
I think I have it solved, but the answer doesn't align with A B C or D.
F_upward =
F_downward =
40 N
F_downward =
F_gravity_down + F_kinetic_friction =
(75N * sin(30)) + F_kinetic_friction =
37.5 N + F_kinetic_friction =
40 N
F_kinetic_friction =
2.5 N
Book solution:
C (because 40N * cos(30))
A back-packer grabs her 75 N pack and drags it with constant speed at an angle of 30 degrees to the horizontal. The force she applies to the pack is 40 N. What is the magnitude of the kinetic frictional force that is acting on the pack?
A. 20 N
B. 28 N
C. 35 N
D. 40 N
My solution:
I think I have it solved, but the answer doesn't align with A B C or D.
F_upward =
F_downward =
40 N
F_downward =
F_gravity_down + F_kinetic_friction =
(75N * sin(30)) + F_kinetic_friction =
37.5 N + F_kinetic_friction =
40 N
F_kinetic_friction =
2.5 N
Book solution:
C (because 40N * cos(30))
No, C is right... you just need to worry about the force on the x-axis. Remember too that the thing moves in a constant velocity, so the net force for all axis is zero...
The only force you got for the x-axis is the x-component of the pulling force (40Ncos30) and the frictional force. So it's given that the force should be equal to 40Ncos30 (34.64 N) and going to the OPPOSITE DIRECTION...
TyphoonStruggle
Member
DEATH™;61797710 said:
Oh. I misunderstood the questions. I thought by 30 degrees to the horizontal, it meant up a ramp of 30 degrees, but when thinking about it she's probably pulling the bag across the ground with a strap angled at 30 degrees.
Thanks.
reposting for new page,my apologies.
math questionszzz!!!
so i'm looking at a graph. its a line curving downwards. i think its an exponential curve. the y intercept is 9. it goes through the points 1,6 and 2,4.
how do i find an equation of the graph?
it's going to be in the form y=a(b)^x, right?
math questionszzz!!!
so i'm looking at a graph. its a line curving downwards. i think its an exponential curve. the y intercept is 9. it goes through the points 1,6 and 2,4.
how do i find an equation of the graph?
it's going to be in the form y=a(b)^x, right?
Not necessary. More information is needed.
so the question is impossibru!?
What's the original question? There are an infinite number of functions can be "curving downwards. the y intercept is 9. it goes through the points 1,6 and 2,4."
oh okay, one second .....
blah, it wont let me save a picture of the graph to post on here (perhaps to prevent me from getting help from people on the internet HA HA.) ..
the original question is
"Find an equation of the exponential curve sketched in the graph. If necessary, round to two decimal places. [Hint: Choose two points whose coordinates appear to be integers.]"
the best i can describe the graph is that its a little blue line. it is decreasing. its y intercept is 9. and according to the HINT, i found two points whose coordinates appear to be integers: 2,4 and 1,6.
the original question is
"Find an equation of the exponential curve sketched in the graph. If necessary, round to two decimal places. [Hint: Choose two points whose coordinates appear to be integers.]"
the best i can describe the graph is that its a little blue line. it is decreasing. its y intercept is 9. and according to the HINT, i found two points whose coordinates appear to be integers: 2,4 and 1,6.
assuming y=a(b^x)
x=0, y=9 -> a=9
x=1, y=6 -> b=2/3
we have, y=9(2/3)^x. Verified with x=2, y=4.
assuming y=a(b^x)
x=0, y=9 -> 9=a(b^0) -> 9=a, since b^0 is always 1, (unless b=0, which 0^0 is undefined.).
Anticitizen One
Banned
Need some Precalc help
Question says Verify the identity. Show all steps:
cosX-sinX/cosx + sinx-cosx/sinx = 2-secXcscX
Other question says Find the exact value by using a sum or difference identity:
cos285 degrees
Question says Verify the identity. Show all steps:
cosX-sinX/cosx + sinx-cosx/sinx = 2-secXcscX
Other question says Find the exact value by using a sum or difference identity:
cos285 degrees
cpp_is_king
Member
Did you forget to include parentheses in the first equation?
Anticitizen One
Banned
cpp_is_king
Member
Err.. i think you got them wrong there too. it should be
(cos(x) - sin(x))/cos(x) + (sin(x) - cos(x))/sin(x)
Try doing it yourself with these parentheses
I'd imagine you mean:
((cosx-sinx)/cosx) + ((sinx-cosx)/sinx) = 2-secxcscx
?
edit: beaten
Anticitizen One
Banned
right sorry. I'm just confused how to solve the problem
In problems like this, a few things that can *sometimes* (not always, but on occasion) illuminate the path are to:
1-Break up weird fractions, like so:
((cos-sin)/cos) + ((sin-cos)/sin) = 2 - sec*csc
1 - sin/cos +1 - cos/sin = 2 - sec*csc
2-Simply types of terms by converting csc and sec to sin and cos, like so:
1 - sin/cos +1 - cos/sin = 2 - sec*csc
1 - sin/cos +1 - cos/sin = 2 - 1/(sin*cos)
Can you see where to take it from here?
Remember, you're just trying to use algebraic manipulations until both sides to look the same.
thequickandthedead
Member
Guys, how do i know where loci1 (|z|=|z-4i|) and loci2 (arg z = 1/6pi) intersect?
cpp_is_king
Member
Something to the negative n is the same as the reciprocal to the n. What you've written is the same as:
sum[ (1 / (3x))^n ]
Use the entire thing, 1/3x as the item that's being raised to the power n
But it's 3x^-n, not (3x)^-n, so the coeficient is not raised.
I think it might not matter to the formula I was trying to use anyway, I think I messed up somewhere earlier on this problem.
cpp_is_king
Member
Oh. So just pull the 3 out of the entire sum then.
sum(3x^-n), n=2 to infinity
= 3 * sum(x^-n), n = 2 to infinity
Okay... I've got an answer on this one, but I don't think it's right and unfortunately it's not one of the problems in the book that I can check the solution. I believe I may have an issue with one of my signs swapping in the problem somewhere.
In this problem, t = a constant.
In this problem, t = a constant.
Hypno Funk
Member
Actually, you have gone a step too far, you have found the solution in your fourth from last line, notice that all the integration is on the left hand side? All you need to do is divide by 2!
Here is my working - I got the same result as yourself.
A quick check, by differentiating the right hand side shown above (using the product/chain rule) does indeed confirm that this is the right answer (ignoring any constants).
Actually, you have gone a step too far, you have found the solution in your fourth from last line, notice that all the integration is on the left hand side? All you need to do is divide by 2!
Here is my working - I got the same result as yourself.
A quick check, by differentiating the right hand side shown above (using the product/chain rule) does indeed confirm that this is the right answer (ignoring any constants).
Thanks for the help and the verification here.
Am I missing a reason as to why the bounds aren't completed for the s values between 0 and t though?
Anticitizen One
Banned
Stuck on another problem:
The problem asks: Find the exact value under the given conditions:
sin a(alpha) = (3/5), 0 < a < (pi/2); cos b(beta) = (20/29), 0<B<(pi/2) Find tan (a+b)
Here is how I did it:
I used the formula tan (a+b) = ((sin(a+b))/(cos(a+b)))
I then used the pyhtagorean theorm to find sin b (I got 21/29) and cos a (I got 4/5)
so plugging them in I get (sin(3/5)+(21/29))/(cos(4/5)+(20/29)) = (192/145)/(216/145)
I cancel out the 145 and I reduce the (192/145) by dividing by 8 to get (24/27)
However the answer key says the correct answer is 144/17
So i'm not sure what i'm doing wrong. I also don't get what 0 < a < (pi/2) and 0<B<(pi/2) even means
The problem asks: Find the exact value under the given conditions:
sin a(alpha) = (3/5), 0 < a < (pi/2); cos b(beta) = (20/29), 0<B<(pi/2) Find tan (a+b)
Here is how I did it:
I used the formula tan (a+b) = ((sin(a+b))/(cos(a+b)))
I then used the pyhtagorean theorm to find sin b (I got 21/29) and cos a (I got 4/5)
so plugging them in I get (sin(3/5)+(21/29))/(cos(4/5)+(20/29)) = (192/145)/(216/145)
I cancel out the 145 and I reduce the (192/145) by dividing by 8 to get (24/27)
However the answer key says the correct answer is 144/17
So i'm not sure what i'm doing wrong. I also don't get what 0 < a < (pi/2) and 0<B<(pi/2) even means
Hypno Funk
Member
My apologies - you are correct. I had only gone as far as the general integration form, and had not evaluated between the the values of s between 0 and t.
By differentiating the right hand side of my final line, and evaluating between 0 and t I do in fact get the same answer as yourself in your last line of working.
You were correct all along.
Sorry for any confusion.
Thanks again for your help and verification. Integration by parts is making my head spin a bit. I'm having a bit of difficulty really feeling like I'm grasping it. Edit: Especially when you start throwing in additional substitutions.
Hypno Funk
Member
I think you might be using the trigonometric identities wrong - but it is hard to tell from your original post - typing math is never easy!
Here is the solution I got, which agrees with the model solution.
The bounds given for a and b being less than pi/2 is just so the value of Tan (a + b) is bounded. at Tan (pi/2) the function tends towards infinity. Tan is not continuous for all x between 0 and pi.
Anticitizen One
Banned
I think you might be using the trigonometric identities wrong - but it is hard to tell from your original post - typing math is never easy!
Here is the solution I got, which agrees with the model solution.
The bounds given for a and b being less than pi/2 is just so the value of Tan (a + b) is bounded. at Tan (pi/2) the function tends towards infinity. Tan is not continuous for all x between 0 and pi.
Thank you so much. Did you type that up or did you use some kind of website?
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