Please help me with this exercise of calculus...?
The Volume of a cube is found by the formula V = x^3, and the surface area of a cube SA = 6x^2 starting with the volume, differentiate V = x^3....V = 3x^2dx (where dx = 0.1) then plug in numbers...V = 3*(30^2)*0.1 = 270 cm^3. To find the relative error, the formula is (I Vmax - V I) / V, which simply means the absolute value of the maximum error value(30.1^3) minus the accepted value (30^3) ...divided by the accepted value(30^3)....I30.1^3- 30^3I = 270.901 / 30^3 = 0.001003337.... that is your relative error, the percentage error is found by multiplying the relative error by 100. 0.001003337 * 100 = 0.1003337%. The surface area is SA = 6x^2. first differentiate, SA = 12x*dx. then plug in numbers, SA = 12*30*0.1 = 36 cm^2. relative error follws the same formula as before, I 6*30.1^2 - 6*30^2 I = 36.06 / 6*30^2 = 0.006677778.... percentage error is Er*100, 0.006677778*100 = 0.667777778%
What are the most difficult exercises of calculus?
If you mean the questions in elementary calculus (differential and integral) that require the most complex calculations, that’s clear: partial fraction expansions of degree more than 4. Technically part of calculus, they’re frequently de-emphasized or omitted altogether since everyone from the instructor to the students recognize them for difficult tedious calculations.At the other extreme, as noted already by another respondent, practical questions (related rates, optimization, etc) often present the student with very nasty difficulties, at least until they remember/learn something (not much!) of allied disciplines, concerning how to map external concepts onto the language of calculus, e.g. now differentiation creates the velocity function from position, and likewise how integration reverses the process.
I need help with this calculus exercise?
. ( a ) Q(t) = 78/( 5 + 72*e^-t ) = lim t→∞ { 78/( 5 + 72*e^-t ) } = lim t→∞ { 78/( 5 + 72/e^t ) } = 78/( 5 + 0 ) = 15.6 thousand = 15600 people ━━━━━━━ ( b ) Q(t) = 78/( 5 + 72*e^-t ) Q(t) = 78 ( 5 + 72*e^-t )⁻¹ Q’(t) = -(1)(78) ( 5 + 72*e^-t )⁻² * (-72e^-t ) Q’(t) = 5616 e^-t / ( 5 + 72*e^-t )² Q’(t) = 5616 e^t / ( 72 + 5*e^t )² ━━━━━━━━━━━━━━ Q’(0) = 5616 e⁰ / ( 72 + 5*e⁰ )² Q’(0) = 5616 / 5929 thousand people / day Q’(0) = 947 people / day ━━━━━━━━━━━ ( c ) Q’(t) = 5616 e^t / ( 72 + 5*e^t )² Q’(5) = 5616 e⁵ / ( 72 + 5*e⁵ )² thousand people / day Q’(5) = 1258 people / day ━━━━━━━━━━━━ ( d ) Q’(t) = 5616 e^t / ( 72 + 5*e^t )² ( 72 + 5*e^t )² Q’(t) = 5616 e^t 2( 72 + 5*e^t ) ( 5e^t ) Q’(t) + ( 72 + 5*e^t )² Q'’(t) = 5616 e^t 2( 72 + 5*e^t ) ( 5e^t ) Q’(t) + ( 72 + 5*e^t )² * 0 = 5616 e^t 2( 72 + 5*e^t ) ( 5e^t ) Q’(t) = 5616 e^t 2( 72 + 5*e^t ) ( 5e^t ) * 5616 e^t / ( 72 + 5*e^t )² = 5616 e^t 56160e^t = 5616 ( 72 + 5*e^t ) 56160e^t = 404352 + 28080e^t e^t = 14.4 t = 2.67 days ━━━━━━
Calculus Help Needed Urgently Please?
For even multiples of 30 minutes, you can find the number of bacteria by using 300*2^n where n is the number of 30 minute intervals that have passed. So: a) 300*2^6 = 300*64 = 19200 b) 300*2^(2t) For times that are not an even multiple of 30 minutes we can write an expression in the form: b(t) = 300e^(kt) where t is time in hours and solve for k 600 = 300e^(0.5k) or e^(0.5k) = 2, so 0.5k = ln(2), k = ln(2)/0.5 = 1.38629436 Therefore c) b(0.67) = 300e^(0.67*1.38629436) = 759.453956 d) 50000 = 300e^(1.38629436t) e^(1.38629436t) = 50000/300 = 500/3 1.38629436t = ln(500/3) t = ln(500/3)/1.38629436 = 3.69 hours
Calculus 2/Physics Question?
The volume of a cone is 1/3 π r^2 h. In this problem, r = h/2, so we can write the volume as πh^3 / 12. Since h = 8, the volume is 128π/3, and the half-tank of oil has a volume of 64π/3, and a weight of 1216π. Working backward, h = ∛(12V/π), so the height of the oil is ∛256 = 4∛4 or 2^(8/3). That's about 6 1/3 feet. Define x to be the distance from the outlet to the surface of the oil, with positive x pointing downward. Before pumping begins, x = 9 - 4∛4. When pumping is finished, x = 9. Those two values will be the integration limits. Notice that x + h = 9, where h is the height of the oil surface. The area of the oil surface is πr^2 where r = h/2 and h = 9 - x, so A = π/4 (9-x)^2. The differential element of volume is π/4 (9-x)^2 dx. Work is force times distance, so put the density in there and integrate: 57 ∫ x dV = 57π/4 ∫(9x - 18x^2 + x^3) dx = 57π/4 (9/2 x^2 - 6x^3 + 1/4 x^4) The limits are from 9 - 4∛4 to 9. When you plug in those limits, you'll have your answer. [Suggestion: From looking at those limits, I think you may get a lot of cancellation if you keep the lower limit as a binomial.]
I need help with this calculus physics application...please answer!! Thanks?
a) i'm just gonna calculate the time it takes to reach the peak, and multiply that time by two, because the time going up will equal the time going down. (v0 = initial velocity BTW) v will be set to zero, because the only time velocity is 0 is when the projectile reaches its peak, which is what we want v = v0 + at 0 = 160 + -32t -160 = -32t t = 5s takes 5s to go up. multiply by two, it will hit the ground in 10 seconds b) we won't need time for this. just this equation: v^2 = v0^2 + 2ax we'll attack this in a similar way as the previous one. make v zero, because that is the peak, and solve for x to give us the peak height. 0 = (160)^2 + 2(-32)x -25,600 = -64x x = 400. it will go 400 feet high. c) here, we plug in 36ft for x, 0 for v again, and solve for v0: v^2 = v0^2 + 2ax 0 = v0^2 + 2(-32)(36) 2304 = v0^2 v0 = 48 ft/s
Where do I find good mathematics exercises from pre-algebra to calculus on the internet?
One good source is Page on chiprime.com-the adaptive quiz there. It starts out easy, but don't let that fool you. Questions get quite challenging as you answer more of them correctly. Once you've worked through these, you'll be well on your way to taking on harder material. Good luck!
What makes calculus so difficult for most people?
Algebra and Trigonometry. Because most people don't have the necessary decent Algebra skills, also some Trigonometry skills are required in order to do well in Calculus. Calculus is actually not difficult at all if you're good at Algebra and Trigonometry. It's just a bunch of new formulas and rules that you have to memorize and you need to apply the Algebra skills when you're solving Calculus problems. And this is why Algebra is so important in mathematics because lots of advanced math courses rely on Algebra. Students should really master Algebra in order to master Calculus or in any other math courses beyond Calculus in college.
What are the most difficult things to learn in calculus?
This really depends entirely on the individual. It could be the jump from single-valued functions to multivariate functions. It could be the shift from advanced integration to sequences and series. Parametric/Polar Equations. Greens/Stokes/Divergence Theorem. There are plenty of things that could be considered hard to some but easier to others, and I've seen this happen over and over.That being said, in hindsight, Calculus isn't really that bad compared to Analysis (rigorously proving everything you take for granted) , in my opinion. And that can still be an opinion: one of my friends was pretty bad at calculus for the sole reason that he thought the presentation was dry (he hates computing things in general), but did a lot better when it was more rigorous and proof-based. So take that with a grain a salt, too.