What is the curvature of the curve y ln x 2 1 at 0,0

Solution: y'=1/x，y"=-1/x²

So the curve is curvature at (1,0).

k = 1 x (1 + 1 x ) 3 2 = 2 4 radius of curvature r = 2 2

Since the curve y=lnx has a tangent slope y at (1,0).'=1, so.

The normal equation y=-x+1, let (a,b) be the center of the circle of curvature, then.

b=-a+1

Again(a-1) +b-0) =8, the solution is a=3, b=-2, so the circular equation of curvature is.

x-3）²+y+2）²=8

It is known that f(x)=e is x+m to the power x, g(x)=ln(x+1)+21If the tangent slope of the curve y=f(x) at the point (0, f(0)) is 1, find the value of the real number m.

2.When m 1, prove f(x) g(x) x (1) f'(x)=e^(x+m)-3x^2

f'(0)=e^m=1

m=02) m>=1

h(x)=e^(x+m)-ln(x+1)

e me x[1-ln(x+1) e x] only need to be proved.

1-ln(x+1) e x>0.

i.e. e x>ln(x+1).

At this point h(x)>0

Hence f(x) g(x) x

Summary. Hello, I am very happy to solve the problem for you, the curvature of the curve y=x 2-sinx at the point (0,0) is solved as follows: dx dy=2y

d²x/dy²=2

So, the curvature is.

k=(d²x/dy²)/1+(dx/dy)]³

2/√(1+2y)³

At (0,0), the curvature is .

k=2, you can refer to it, if you have any questions, I am answering <> for you

The curvature of the curve y=x 2-sinx at the point (0,0) is.

Hello, I am a mentor who loves to think and answer, please wait a while, I will answer you immediately after sorting out the questions here, thank you for your understanding

Is it 0. Hello, I am very happy to answer for you, the curvature of the curve y=x 2-sinx at the point (0,0) The solution idea is as follows: no reed dx dy=2yd x dy=2So, the curvature is k=(d x dy) 1+(dx dy)] 2 (1+2y) At (0,0), the curve of Qu Kuchang with the rate is fast and clear k=2, you can refer to Ha, what questions I am answering for you<>

dx/dy=2y？？？What a mess.

If y≠0, then dy y=2x, so, ln|y|=x +c0, y= e c0·e (x) = c·e (x) Since c= e c0≠ Pi Huifan 0, the general solution does not contain a solution with over-burning hail (0,0). To sum up, Biling has and only one solution of (0,0): y=0

y'=1/x,y"=-1/x?So the curvature of the curved profile at (1,0) k=1 x?/（1+1/x?

3 2 = 2 4 radius of curvature r = 2 2 due to the curve y = lnx at (leakage cavity 1,0) tangential slope y'=1, so the normal equation y=-x+1, let (a,b) be the center of the curvature circle, then b=-a+1 and (random search and guess a-1)?+b-0）?=8, the solution is a=3, b=-2....

Summary. Hello dear, I'm glad to answer your <>

Since: y(x)=2x-3x2, y(x)=2-6x, so: y(1)=-1, y(1)=-4, so the curvature of the curve y=x2(1-x) at the point (1,0):

k=|?4|(1+1)32=2 So the answer is: 2

The curvature of the curve y=x2-1 at the point x=1 2.

Hello, dear, very high orange shouting for you to answer <>

Because: y (x) = 2x-3x2, y (x) = 2-6x, volt so: y (1) = -1, y (1) = -4, so the curvature of the curve y = x2 (1-x) at the point (1, missing panicle 0):

k=|?4|(1+1)32=2 So the answer is: 2

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Original = (0, 2) x (1+4x square) dx1 8 (0, 2) (1+4x square) d(1+4x square) 1 8 2 mountain with 3 (1+4x square) (3 Zhengwei dou 2)|(0, 2)1 Lifting mill 12 [27-1].

To find the curvature of the curve y=x at the point (1,1), you can do the following steps:

1. Find the first and second derivatives of the function respectively, that is, the wax stool.

y'=3x²

y"=6x2, find the curvature in the Cartesian coordinate system.

3. The curvature of the wild roll of the curve y=x at the point (1,1).

y=x^3y'=3x^2

y'(1)=3

Curve y=x 3 curvature =y at point (1,1).'Liquid spring stupid (1) = 3

Summary. Kiss <>

We'll be happy to answer your questions<>

1) y=4x-x 2 solution: y'=4-2x，y''=2, substituting the curvature formula k= frac}}, and getting k= frac}}(2) y=sinx solution: y'=cosx，y''=sinx, substituting the curvature formula k=frac}}, we get k=|sinx|

Find the curvature of the following curve at any point: -|1) y=4x-x^2;-|2) y=sinx, kiss <>

I'm glad to answer <> for you

1) y=4x-x 2 solution: y'=4-2x，y''=2, substituting the curvature formula k= frac}}, we get k= frac}}(2) y=sinx:y'=cosx，y''=sinx, substituting the curvature good letter cover formula k=frac}}, we get k=|sinx|

Data Expansion: (1) The first derivative of the y=4x-x 2 curve is y'=4 - The number of the second-order guide circles of the 2x curve is y''2 Therefore the curvature of the curve is |y''|1+y')3 2) =2 [(1+(4-2x) )3 2)](2) y=The first derivative of the sinx curve is y'The derivative of the second-order scaboo of the =cosx curve is y''sinx hence the curvature of the curve is |y''|1+y'²)3/2) =1 / 1+cos²x)^(3/2)

Find the point with the smallest radius of curvature on the curve y=x2-2x, and find the center coordinates of the corresponding circle of curvature.

Kiss <>

We'll be happy to answer your questions<>

The radius of curvature of the curve y=x -2x is r=|1+4x²|^3/2)/|2x-2|。In order to find the point with the smallest radius of curvature on the curve, we need to find the derivative of the radius of curvature and make it 0: dr dx = 4x(2x-1)) 2x-2|^5/2 - 8x(1+4x^2))/2x-2|3 2 = 0 Simplification yields:

x(4x-3) =1 is solved: x = 1 2 or 3 4 where when x = -1 2, the radius of curvature is the smallest, and the radius of curvature is r=1 2, that is, the radius of the corresponding circle of curvature is 1 2. To require the central coordinates of the corresponding curvature circle, we need to find the tangent equation of the point on the curve, and then find the coordinates of the intersection of the tangent and the x-axis.

The slope of the curve y=x at the point (-1 2, 5 4) is y'=2x-2 = 3, so the tangent equation for this point is y = 3(x+1 2)+5 4. The equation intersects with the x-axis, and the coordinates of the intersection point are (-5 6, 0). Therefore, the central coordinates of the corresponding circle of curvature are (-5 6, 1 2).

Summary. "Limit" is a fundamental concept of calculus, a branch of mathematics, and "limit" in a broad sense means "infinitely close and never reachable". "Limit" in mathematics refers to:

In the process of a certain variable in a function, which gradually approaches a certain definite value a and "can never coincide to a" ("can never be equal to a, but taking equal to a' is enough to obtain high-precision calculation results"), the change of this variable is artificially defined as "always approaching without stopping", and it has a "tendency to constantly get extremely close to point a". Limit is a description of a "state of change". The value a that this variable is always approaching is called the "limit value" (which can also be represented by other symbols).

2.The level of the curve y=1 ((4x 2+x))ln(2+1 x).

Comrade, you can just shoot me the question.

2.The curve y=1 ((4x 2+x))ln(2+1 x) is the water of the net Qingping for the tease grip y 0. The specific process is slightly referred to and sent by the teacher**.

Finding the horizontal asymptote is the limit of the approach with an x.

There is a horizontal asymptote.

First of all, calculate the limit problem, one of the three major calculations of high mathematics. Secondly, observe which of the 7 infinitives is determined. Secondly, the use of extreme tools, Lopida, equivalent, Taylor, identity deformation, rationalization and other methods.

Finally, according to the four algorithms, the answer can be obtained.

"Limit" is a fundamental concept of calculus, a branch of mathematics, and "limit" in a broad sense means "infinitely close and never reachable". The "limit" in mathematics refers to a certain variable in a function, which gradually approaches a certain definite value a in the process of eternal change of cavity impulse (or decreases) and "can never coincide to a" ("can never be equal to a, but taking equal to a" is enough to obtain high-precision calculation results), and the change of this variable is artificially defined as "always approaching without stopping", and it has a "tendency to constantly get extremely close to point a".

Limit is a description of a "state of change". The value a that this variable is always approaching is called the "limit value" of the trapped round silver (of course, it can also be represented by other symbols Wang Yan).

How did 0 come about?

Can you go into more detail about the process?

The denominator is approaching.

The denominator is infinity and the fraction is 0

This limit is too simple, and the teacher wrote it in great detail.

Generally, this kind of question is done in one step.