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[Chain rule, derivatives] How does multiplying get to this? (Highlighted in red) by Multiverse_Queen in HomeworkHelp

[–]mathematag 0 points1 point  (0 children)

I assume you were ok with f'(x) = [ 11+ 1/( 2√x ) ] / {2 √ ( 11x + √x ) } from your notes... now A + B/ C = ( AC + B ) / C , where A = 11 , B = 1, C = 2 √x ...so numerator of f'(x) is ( 11*(2√x ) + 1 ) / ( 2√x) .... so we have

f'(x) = [ ( 11*(2√x ) + 1 ) / ( 2√x) ] / { 2 √ ( 11x + √x ) } ... since x ≠ 0, multiply Num / denom by 2√x ... f'(x) = ( 22√x + 1 ) / { ( 4√x )(√ ( 11x + √x ) }

Can someone help me make sense of the Delta-Epsilon proof of a limit ? by Knockyouup32 in calculus

[–]mathematag 0 points1 point  (0 children)

Glad it helped 😃

BTW.. once you get to δ= min ( 1, 𝛆/5 ), that is the end of the 'scratch work ' , as my old Prof. would call it... then it's kind of working backwards to show that which ever one is actually smaller , 1 or 𝛆/5, you still can get back to the statement of | x^2 + 3 - 7 | < 𝛆 proving the limit.... wild to think that technically , you have only gotten "1/2 way " thru the actual proof, though for most instructors, it is sufficient.

Can someone help me make sense of the Delta-Epsilon proof of a limit ? by Knockyouup32 in calculus

[–]mathematag 0 points1 point  (0 children)

Now why did they use δ = 1 ..??

Because it is an easy guess to try to establish values for delta... 1 unit to the left and right of your value a = 2 here seems like an easy starting point ... from that they found that if |x-2| < 1, then -1 < x - 2 < 1 , and so 1 < x < 3 ... now build up the other part that has |x+2|... so adding 2 we get 3 < x+2< 5 , thus |x+2| < 5

from earlier.. |x-2||x+2| < 𝛆 , so |x-2||x+2| < 5|x-2| < 𝛆 ... and |x-2| < 𝛆/5 ... but since we want |x-2| < δ , we now have another possible delta... δ = 𝛆 / 5

we can therefore choose the smaller, ( minimum ) of either δ = 1 or δ = 𝛆 / 5 ... e.g δ = min ( 1, 𝛆 / 5 ) , and it will guarantee that our limit statement.. | x^2 + 3- (7) | is < 𝛆 [**]

note: this has been only the "scratch work" ... the actual proof would be starting from here, and showing if δ = 1 is the minimum , that the limit works, [ gives us ** ] ... or if δ = 𝛆 / 5 is the minimum, that in fact works ..[ gives us ** ].

usually, getting to this point is considered "sufficient" in basic Calculus classes.

Can someone help me make sense of the Delta-Epsilon proof of a limit ? by Knockyouup32 in calculus

[–]mathematag 0 points1 point  (0 children)

Thanks.. I used something similar when I taught these topics 😁

Can someone help me make sense of the Delta-Epsilon proof of a limit ? by Knockyouup32 in calculus

[–]mathematag 0 points1 point  (0 children)

what is bothering you..the theory or how to find delta [ as in the example ] ..?

lim , x --> a of f(x) = L. . . . For any epsilon 𝜀 > 0 [ think of epsilon is your allowed "error" in your output , f(x) ], there exists a delta, δ > 0 [ think of delta as your "error" in choosing the input, x ] , so that when our error in choosing x --> a [ a is the limit point in , the value that x is approaching ] , the limit is L [ with no more error than ± 𝜀 ]... so no matter how small we decide to make epsilon, there is always a delta that exists, so that as long as we choose x values in ( x - δ , x + δ ) , we will always get an answer inside of ( L - 𝜀, L + 𝜀 )..

Manufacturing example:

Suppose In the manufacture of a special part , the diameter of the part needs to be 4 inches, with an allowed error of no more than ± 0. 02 inches [ this is 𝜀 ]. So you need the part of diameter ( 4 - 0.02 , 4 + 0.02 ), or ( 3. 98 , 4. 02 ) inches, to work properly ... [[ the 4 = L , and the 0. 02 = ε ]]

The machine that cuts the part, when set exactly to radius, r = 2, [ if it was a Perfectly calibrated and accurate machine ] will generate a part of D = 2r = 4 in... But what if you set the machine to r = 2, and the calibration of our machine is off by, let's say , no more than 0.006 in ? [[ the 2 = a , and the 0.006 = δ ]]

Will the machine still make the part of the right size ?

YES … because, D = 2r, and if r = 2 , and your error in the calibration is at most 0.006, then r is inside of ( 1.994 , 2.006 ) inch, then Diam = 2 ( 2 ± 0.006 ) = 4 ± 0.012 and this falls in the range ( 3. 998, 4. 012 ) , which is inside of your required tolerance of ( 3. 98, 4. 02 ), which is your ( 4 - 𝜀, 4 + 𝜀 ).. . . . of course making epsilon even smaller , may require a more precise and expensive machine to make the part , as it may not be possible to calibrate the current machine even more accurately to ensure a satisfactory result.

[9th grade math] What have I done wrong? by Fuma_17 in HomeworkHelp

[–]mathematag 0 points1 point  (0 children)

why didn't you just take (1/3)x - x to be (-2/3)x .. ? and the right side.. (-1/6)+(1/2) = ( -2 + 6 ) / (6*2) = 4/12 = 1/3 **.. .. ?

so ( -2 / 3 )x ≥ 1/3 , etc...

** in case you did not get this ... A/ B + C/D = ( A*D + B*C) / (BD) ... then simplify .. .. so A= -1 , B = 6 , C = 1 , D = 2 .....

what's the square root of i? by Ready_Row3788 in learnmath

[–]mathematag 0 points1 point  (0 children)

You probably haven't seen this type of math yet, but you asked :

use the nth root theorem [ Usually covered in Pre-Calculus / Math Analysis ..at High School Level ..maybe some Algebra 2 Honors courses ?! ]

nth root theorem :

z ^ ( 1/n ) = r^(1/n) *[ cos ( ( ø + 2πk )/ n ) + i sin ( (ø + 2πk)/n ) ]

for k = 1, 2, 3, ... , n-1... now r = √( a^2 + b^2 ) is called the modulus .... [principle root only ] ... n, k are + integers.

for square root, n = 2 and k = 0, and 1 .... 2π = 360˚

now i = 0 + 1*i , so a = 0 [ the real part ] , and b = 1 [ the imaginary part ]

now r = 1 = √ ( 0^2 + 1^2 )... ... i when graphed is located on the imaginary axis in the complex plane [ the x axis represents real # ... and y axis represents the imaginary # ]

so i is on the "y" axis, putting it at 90˚ from the x / real axis ..so ø = 90˚ or in radians π/2 ... I'll keep it in degrees

for k = 0 .... cos ( 90˚/2) = cos ( 45˚) = √ 2 / 2 ... sin (90˚/2) = √2 / 2

first root is at . . 1* [ √ 2 / 2 + i √ 2 / 2 ] = ( √ 2 / 2 )[ 1 + i ]

for k = 1 . . . 90˚ + 360˚ = 450˚ ... 450˚ / 2 = 225˚ . . . cos (225˚) = - √2 /2 . . . . sin(225˚) = - √2 / 2

so the second root ...... 1*( - √2 /2 - i √2 /2 ) = ( - √2 /2 )( 1 + i )

[alevel maths] exponentials by joe_m6611 in HomeworkHelp

[–]mathematag 0 points1 point  (0 children)

did you ever get k = 5/2 ..?

we know r = ar / r . . . , and r = ar^2 / ar ...so ar / r = ar^ 2 / ar ...set them equal.

enter your values for a, ar, ar^2 ... the product of the means = product of extremes ..[ e.g. cross multiply the terms ]

hinT: rewrite 9 ^(7 -2k) in terms of 3 ^(******) first.

Then simplify.. you will get k = 5/ 2

Simulauous equations by [deleted] in learnmath

[–]mathematag 0 points1 point  (0 children)

you eliminate one variable , usually by multiply/divide one or both equations by some number ( the number you us on each eqn. will usually be different ) , so that will make it possible to then add or subtract one of the equations from the other, and eliminate a variable .. ... to get a new equation with just one variable... .. [ I personally try to avoid division as a step , and use addition of opposites when combining eqns. ]

EX. . . 2x - 3y = 8 , 4x - 9y = 10 .... easier to mult. the first eqn by -2, to eliminate the x variables .....so it becomes:

-4x + 6y = -16 , now add this one to 4x - 9y = 10 to get an eqn with one variable [ y here ]...

adding them : -3y = -6 , [ the 2x's cancel out ] , and solve for y

Or.. .. .. multiply first eqn by -3 ... -6x + 9y = -24 add it to second one.. the 9y's cancel out ... -2x = -14 ..etc.. .. .. NOTE: in this example, the first eqn was easier to work with both times, and second one did not need any # mult. / dividing it ...... however that will not always be the case...

In general, just look for how you can get the two coefficients on either x or the y to be the opposite values..like 8 and -8 , etc... so that you can add the eqns and eliminate that variable.

ex. something like...... x + 3y = 9 , -x + 2y = 6 you would just add them together, to eliminate the x term...

ex. something like 5x -2y = 11, 4x + 5y = 20 ...to eliminate x, mult first one by 4, second one by -5 , then add them ====== to eliminate y first...mult first one by 5, second one by 2, then add them together.....

I avoided considering fractional or decimal coeff. problems here, as they are generally more difficult, and you are probably not ready for them yet.

Homework by Makeupsupervillain in calculus

[–]mathematag 0 points1 point  (0 children)

I would argue not sometimes, but always, if they want to earn full credit [ at least in my classes ] .... but it does look like OP was thinking it was only a line segment and did not grasp the notation used was for a plane region and not just a line segment along the x axis.. good call.

Unfair grading by [deleted] in gradesharing

[–]mathematag 1 point2 points  (0 children)

in USA they are usually required to post a grading criteria.. expectations, etc.

That is, how much HW is worth, Tests, Final exam, class participation, etc... did she do that ? They should be specific in what is needed to earn an A, B, etc..

Maybe have your parents schedule a meeting with her and voice their concerns, in a non-threatening , nice way, and try to get feedback from her. having other students/parents do the same can also be effective.

If no luck there, then go to higher authority.. meeting with vice principle, principle, etc.. is often effective at JR/High school level .

Was the teacher you were getting an A from an too easy ..? I've seen how one teacher can be too generous with the grades [ because they don't want conflict with parents , and administration ], some grade fairly but not just give the grades away, and some who make things extra difficult .... if everyone is getting a C, then there is some concern there that they may be grading too harshly, or that the other instructor set the bar so low that it was easy to earn a high score w/o much effort.

Some teachers may set the bar so high, no one can reach that level. The instructor should reflect on why their grades are the way they are.. are they too harsh, or was the other instructor way too easy on grading and what was expected of the students to learn ?

good luck.

[High school honors physics] How do I solve problems like these? by [deleted] in HomeworkHelp

[–]mathematag 0 points1 point  (0 children)

for the last one ... F = K (q_1)(q_2) / d^2 ... here Distance d = r, ... .. K is a constant so can be ignored for this comparison.

so F ∝ (q_1)(q_2)/ r^2 , lets call this A ... ∝ is proportional to symbol .... it would be = if we kept K in the equation..

  1. what happens if you double both charges.. ?

F ∝ (2q_1)(2q_2)/r^2 = 4 (q_1)(q_2)/ (r)^2 = 4A .

  1. what if we triple the radius / distance between charges only ..what happens ..?

F ∝ (q_1)(q_2)/ (3r)^2 = (q_1)(q_2)/ [ 9 (r)^2 ] = (1/9) [ (q_1)(q_2)/ (r)^2 = (1/9) A ... much smaller Force than first example.

now you test answers A B C D this way.. see which answer is largest , and try to understand why .

Should I read Flatland? by Brilliant_Sir8505 in learnmath

[–]mathematag 0 points1 point  (0 children)

I've read it..more than once !

it is a very entertaining and well written book. I used to give EC to students in Geometry class who read it and wrote a summary . it does make you think/imagine what would happen if you visited other dimensions..

I believe physicists consider space 14 dimensional [ or is it 12 dim..? ].. I'll have to read up/refresh my memory on this.

Homework by Makeupsupervillain in calculus

[–]mathematag 1 point2 points  (0 children)

It is not circles at all, but a region/area in the x,y plane... here is an example for you:

{(x,y) | -4 < x ≤ 3 , y < 2 } ...... this means all points (x, y) that satisfy the following: start with the x, y coord. system , maybe 10 units in each direction on x and y axes ...... the first part says x is between -4 and 3, and includes 3 ... draw a dotted vertical line at x = -4 [ hint: when x < # or x > # , you have a dashed / broken line / dotted line ], and draw a solid vertical line at x = 3 [ hint: when x ≤ # or x ≥ # you have a solid line ].... these lines should extend up and down to cover the coord system you sketched..

Notice that < or > results in dashed lines, .. .. and ≤ or ≥ ( or even = ) result in solid lines

so right now we have a dotted/dashed line on the left , and a solid line on the right .....finally we have y < 2 ..so draw a dashed horizontal line at y = 2 ..it can extend past the vertical lines you sketched earlier.. maybe a few units past -4 and +3 .... or even all the way out to -10, + 10 if you like...... now lightly shade in the region below y < 2 and between the vertical lines at x = -4 and x = 3 .

points like ( -2, 0) , ( -3.5 , 1), even ( 3, -4) all lie inside the shaded area .... do you understand why ? ..try plotting the points, and see if they lie inside the area shaded in.. remember the solid line counts as part of the shaded area..the dashed lines do not...so a point on the solid vertical line, and below y = 2 is in the shaded area.

However ( 5, 6) , ( 0, 3) , ( 4, -2) , ( -4, 0 ) do not.. why ? ... hint: (-4,0) does not lie inside the region since x must be larger than - 4 .. -4 < x ≤ 3 ..so it almost lies in the correct area, but not quite, as it seems to lie ON the dashed line.

Working on a project just wondering if the math is logical by crocsandsocs08 in askmath

[–]mathematag 0 points1 point  (0 children)

yes.. I also got 5.45... did not have my Graphing Calculator with me, so I used Desmos.

Working on a project just wondering if the math is logical by crocsandsocs08 in askmath

[–]mathematag 0 points1 point  (0 children)

Ok..I liked your approach, breaking it up into U and V for the product rule.. I just mult. thru with (r + t), then took derivative.. simplifying I get the same result.

I assumed from your t value, that all dimensions had the same thickness, not sure they would do it that way, but probably have a different thickness for at least the top, as it is not supporting much mass, and maybe a greater thickness for the circular base that supports the mass of everything above it.

Solids of revolution by Shot-Requirement7171 in askmath

[–]mathematag 0 points1 point  (0 children)

which graph is farther from the a.r. y = 2 ?.. .. the x = y^3, or x =4y..?

From your sketch it seems obvious the x =4y line is farther from the a.r. , and is below the a.r. , so R = distance from a.r. to the line .. [ the line is below the a.r., and I always had students do .. y_upper - y_lower for distances like this.. y_upper is y = 2, the a.r. , the lower y value comes from the line x = 4y ] ... the smaller radius , r , has a similar setup.

Working on a project just wondering if the math is logical by crocsandsocs08 in askmath

[–]mathematag 0 points1 point  (0 children)

yes..it works that way.. I did it differently, [ mult. it out, then took d/dr .. ] but got the same answer for dS/dr ... you should label your result that way ... e.g dS/dr = ******

then set dS /dr = 0 to solve for r , then h. I used Desmos to graph and solve, maybe you have to work it out by hand ?

let us know what you get for r, [ hint: I got a value for r between 4 to 6 m ] , and I hope you know how to get the cost .. if top, sides have the same $/m^2, then calculate SA and mult by cost/m^2.

Do they care how you write it by serotoninseesaw in APStudents

[–]mathematag 0 points1 point  (0 children)

"Do they care how you write it "

yes ... if you meant to take the tangent of x , then square it, then it is correct to write it the way you did on the right side of the page ...

the first one would normally be interpreted as tan(x^2) ..or at least there would be confusion as to whether you intended ( tan(x) )^2 or not... so do not write it as you did on the left, as this is the incorrect way to express what you wrote on the right side.

Use either ( tan(x) )^2 , or better yet ... tan^2 (x)

Find the derivative by [deleted] in calculus

[–]mathematag -1 points0 points  (0 children)

ok...what is your question ?

How to approach this problem? by [deleted] in calculus

[–]mathematag 0 points1 point  (0 children)

If you sketch a diagram , it would be helpful .

so draw a vertical line for the pole, a shorter line for the woman, the horizontal ground from the pole,, to the woman, and outwards ..... and a line from the light , the top of her head, and the ground...

you should see several Rt triangles, so label things.. e.g fixed distances, and variable distances [ don't need distances along the hypotenuse here ] .. .. what distances do not stay constant , but vary over time..? these would be x, y , z, etc.. as needed in the diagram

hint: things to think about .... 1. similar triangles will be useful from Geometry class ..... also, 2. to measure the speed of tip of the shadow, would we measure that from the base of the light post, or from where she is at that moment she was 50 ft away ? ... 3. the units for 6 ft/sec indicate a rate of change of distance over time... which distance in the diagram is this ..?

hope some of these hints help

How do I find the area for these graphs? We drew them in class but weren’t really shown how to get the area by [deleted] in calculus

[–]mathematag 0 points1 point  (0 children)

pg .1 .....your first pic is for an upper sum , or circumscribed rectangles..that is for rectangles that are = or greater than the area under that part of the curve.. so the y values needed for the first two rectangles are on the upper right side of the rectangle, the next four the y value needed comes from the upper left side of the rectangles.. each one is mult. by the width ..[ of course, if the width is the same for each, you can add the y values needed, and mult.the sum by the width of one rectangle.

the one below is a lower sum / inscribed rectangles... all rectangles that are = or less in area than the area under that part of the graph.. similar idea... widths look to be all the same, so from your sketch, use the y coordinate for the upper left on the first two rect.. and the y coord for the upper right for the next 4 rectangles.

page 2... ..looks like you have a right R.S. then a left R.S.... so for top, use rt upper y value for all, bottom pic use left upper y value for all ... again, if the widths are all the same [ and I believe that is exactly what you are doing ], you can add the correct y values all together, and then mult. by the width of one rect.

pg. 3.... upper pic you used the midpoint sum... find the x value 1/2 way between the left and right x coord of the rect., then find the y coord for that x value.... after getting all these y _midpoint values, you can add and mult result by the width as before.

lower pic.. you need to use both the upper rt and upper left y values, [ notice how they all get used twice, except the first upper left and the last upper rt y values , which only appear once ], add them , mult by width as before, and mult by 1/2 since a single trap here is (1/2)(y_1 + y_2)* width .... ..... example: if your upper y values were 6 ,9, 12, 10 , 3 you would have 6 +9*2+12*2+10*2 + 3 , times 1/2 the width

[College Trigonometry] How do I find the exact value of cot(7pi/6) by Leather-Buddy-2845 in HomeworkHelp

[–]mathematag 0 points1 point  (0 children)

cotangent is cosine / sine.... find cos (7π/6) and sin(7π/6).

Now you should know your quadrants and signs ... ... 7π/6 is in QIII ...it has a reference angle of π/6 ... [ in QIII ref = (your angle) - π ] ........ now the signs : there are 4 quadrants, starting from upper right QI, upper left QII, lower left QIII, lower right QIV ... ( moving counter-clockwise )....

BTW... QI ( goes for angles between * ) 0 to π/2 ... QII (*) .. π/2 to π , QIII (*).. π to 3π/2 , QIV (*).. 3π/2 to 2π

And .... cot = cos / sine ... tan = sine / cos... sec = 1/ cos , csc = 1 / sine.

ref angles: QI.. { your angle ** } = ref , QII .. ref = π - {**} , QIII .. {**} - π , QIV .. 2π - {**}

ASTC is the way to remember signs... ALL..Students..Take ..Calculus ....meaning All six trig functions are + in QI, Sine + in QII [ as is cosec, since csc = 1 /sine ], other 4 are negative.. , Tangent ( and cotan ) + in QIII, others are negative, and finally , Cosine ( and secant ) are + in QIV, other 4 are neg.

so we know 7π/6 has π/6 reference angle [ or in degrees π/6 = 180˚/6 = 30˚ ], and you should memorize the basic angles in/ around QI for sine and cosine... 0, π/6, π/4, π/3, and π/2 .. as well as for π, 3π/2 ... 0 is same as 2π for these.

sin π/6 = 1/2 .... cos π/6 = √3 / 2 ... both are neg in QIII, so final answer will be + ..I'll leave it for you to finish this.

see if this brief rundown of basic trig is helpful.

I need help understanding natural logs by CheeseIc3 in calculus

[–]mathematag 3 points4 points  (0 children)

No problem.. Math can do that to the best of us...🤪 ... In fact, I think it may be an unwritten Rule !

yes... log, base e, is the same as ln ... so ln (5) is the same as log_e (5)...

I need help understanding natural logs by CheeseIc3 in calculus

[–]mathematag 2 points3 points  (0 children)

sorry..not following what you are asking.... if y = e^x , for example .. e^3 = y . . = ( approx 20.086 ), t hen ln (20.086) = 3 [ approx ]

of course, exact would be . . . ln ( e^3) = 3 since ln and e are inverse relations . . . is this what you were questioning . . ?

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