Differentials?
a) Error of the surface area is dA Error of circumference is dC = .5 cm C=2πr, dC = 2π dr dr = dC / 2π = .5/2π Since Area A = 4πr^2 dA = 8 πr dr Since C = 88 cm. C=2πr r=C/2π= 88/2π r=44/π With dA = 8 πr dr dA = 8 π (44/π) (.5/2π)= 28.011 Max error is 28. cm^2 Area = 4πr^2=4π(44/π)^2 =2465 cm^2 Relative error= Max error/Area % 28./ 2465 %=1.136 % b) V = 4/3 π r^3 Error in volume dV = 4 π r^2 dr = 4 π ( 44/π )^2 (.5/2π) =196 ==> Max error in volume Volume = 4/3 πr^3 = 4/3 π (44/π)^3 =11508 cm^3 The relative error= Max error in vol/ Volume =196 / 11508 =1.703 %
How to use differentials to show that...?
This can be done a couple ways, but I don't see the point of using differentials. Is this for an assignment? n does not have to be large. √(n+a) - √n ≅ a / (2√n) Divide both sides by a. [√(n+a) - √n]/a ≅ 1/(2√n) lim a-->0 [√(n+a) - √n]/a = 1/(2√n) Meaning for arbitrarily small values of a, the limit as the secant line becomes tangent (first derivative). You could do a linear approximation, which is what first came to my mind, but it's not necessary.
Why we use differential?
Calculus (differentiation and integration) was developed to improve this understanding.Differentiation and integration can help us solve many types of real-world problems.We use the derivative to determine the maximum and minimum values of particular functions (e.g. cost, strength, amount of material used in a building, profit, loss, etc.).Derivatives are met in many engineering and science problems, especially when modelling the behaviour of moving objects.Our discussion begins with some general applications which we can then apply to specific problems.1. Tangents and Normals which are important in physics (eg forces on a car turning a corner)2. Newton's Method - for those tricky equations that you cannot solve using algebra3. Curvilinear Motion, which shows how to find velocity and acceleration of a body moving in a curve4. Related Rates - where 2 variables are changing over time, and there is a relationship between the variables5. Curve Sketching Using Differentiation, where we begin to learn how to model the behaviour of variables6. More Curve Sketching Using Differentiation7. Applied Maximum and Minimum Problems, which is a vital application of differentiation8. Radius of Curvature, which shows how a curve is almost part of a circle in a local region.You can really relateThanks for A2A
How do I use differentials to estimate [math](8.06)^{2/3}[/math]?
The derivative of a function [math]f(x)[/math] is defined to be[math]f’(x) = \lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h}\tag{1}[/math]Now if [math]h[/math] is small we can write[math]f’(x) \approx \dfrac{f(x+h)-f(x)}{h}\tag{2}[/math]Now from (2) we can write[math]f(x+h) \approx f(x)+hf’(x) \tag{3}[/math]Now if [math]f(x) = x^{2/3}[/math], [math]x=8[/math] and [math]h=0.06[/math] then from (3) we have[math]\begin{align*}8.06^{\frac{2}{3}} &\approx 8^{\frac{2}{3}} + 0.06 \times \dfrac{2}{3} \times 8^{-\frac{1}{3}} \\ &\approx 4 + 0.06 \times \dfrac{2}{3} \times 0.5 \\ &\approx 4.02\tag*{}\end{align*}[/math]The actual value is [math]4.019975083[/math] and so the relative error is [math]6.2 \times 10^{-6}[/math].
Use differentials to estimate the change..?
. step 1: y = 2x² y’(x) = 4x y’(3) = 4(3) = 12 step 2: y’(x) ≈ Δy / Δx Δx = x₁ - x₀ = 3.3 - 3 = 0.3 step 3: Δy = y’(x) * Δx Δy = 12 * 0.3 Δy = 3.6 ━━━━
Use differentials to approximate the change in f(x) = tan x as x increases from pi/6 to pi/4.?
df = sec²x dx Δf ≈ sec²x Δx Δf ≈ sec²(π/6) (π/4 - π/6) Δf ≈ [2/sqrt(3)]² (π/12) Δf ≈ π/9 ≈ 0.35
How can I know what method to use to solve a differential equation by looking at it?
To begin with, any function that is differentiable can be differentiated using the definition of the process. which is:However, seldom is direct application of limit theory an easy approach; but, when coupled with a computer that can perform the iterative approximations, you can always get the answer if there is one. There is a whole branch of mathematics devoted to differential calculus with no other ambition than to invent more short-cut rules for ever more obscure functions in order to avoid using limit theory.So, there is no simple answer to your question without a working knowledge of the fundamental rules for differentiating specific function forms. The trick is to learn the general function forms that are differentiable, then follow the general rule of divide and conquer. That is, we try to rearrange and/or dissect the function into a combination of recognizable differential forms.I strongly recommend some structured learning through a text, local community college, or an on-line course — which there are many. The Great Courses have numerous courses aimed at self-study . 'differential calculus' shows the courses they offer from highly qualified university professors.There are also many free or nearly free on-line offerings. Find one that you like, and take off. Pauls Online Math Notes is an example of one site devoted to assisting enrolled students with their homework.There are numerous fora devoted to mathematics. The Stack Exchange has several for all levels of mathematical endeavors. Mathematics Stack Exchange is the Exchange for general math. Here members spend time getting their hands dirty with real problems.If you have a specific differentiation problem, I may be able to make some suggestions.
Is there some way to use differential geometry to prove that the Earth isn't flat?
Yes, indeed. That’s how the size of the earth was determined in the 1700s. See the Wikipedia articles History of geodesy and especially French Geodesic Mission for a summary of a project to see how close the earth was to a perfect sphere.On the earth, you can measure distance so you can triangulate portions of the earth’s surface, and from the measurements, you can determine curvature of the earth. Curvature is the reciprocal of the radius of the earth.
How do you use differentials to approximate cube roots?
First, lets make a functions for the cubed root of x: y = f(x) = x^(1/3). Next, lets find the derivative, simply using the power rule (d/dx x^a = a * (x^(a - 1))) dy/dx = 1 / (3 x^(2/3)) dy = 1 / (3 x^(2/3)) dx dy is the change in y and dx is the change in x. Now, lets find a number we know the cube root of and is close to 26 (which is the number we want to find the cube root of). 1^3 = 1 2^3 = 8 3^3 = 27 I would say that 3^3 is pretty close. Now, let x = 27, thus the change in x is: 26 - 27 = -1 Now lets plug those values into the differential equation: dy = 1 / (3 x^(2/3)) dx dy = 1 / (3 (27)^(2/3)) (-1) dy = -1 / (3 (3)^2) dy = -1 / (3 * 9) dy = -1 / 27 dy = -0.037037 Now, the answer is: y + dy, at x = 27: y + dy (x^(1/3)) + (-0.037037) (27^(1/3)) - 0.037037 3 - 0.037037 2.96296 Therefore, the approximate cube root of 26 is 2.96296. To then test this out, you can multiply it by it's self three times to come up with: 26.0123, which is not exact, but fairly close, which is what we wanted (an approximate).
Use differentials to estimate the propagated error!!?
The measurements of the base and altitude of a triangle are found to be- base: 22 cm and altitude: 35 cm. The possible error in each measurement is 0.3cm. Use differentials to estimate the propagated error in computing the area of the triangle. Please show all work and round your answer to four decimal places. Please help me with this! Thank you!!! =)