At what point does the curve have maximum curvature? y=9ln⁡(x) What happens to the curvature as x→∞ ?

Question

At what point does the curve have maximum curvature?  y=9ln⁡(x)  What happens to the curvature as x→∞ ?

Pohanginah · Accepted Answer

See the photo below:

Jeffrey Jordon · Accepted Answer

We are giveny=9 in xWe can use formulak(x)=|y″|(1+(y′)2)32y′=9xy″=−9x2now, we can plug it and we getk(x)=9x2[1+(9x)2]32k(x)=9x(x2+81)32now, we can find derivativek′(x)=ddx9x(x2+81)32=9⋅ddx(x)(x2+81)32−ddx((x2+81)32)x((x2+81)32)2=9⋅1⋅(x2+81)32−3xx2+81x((x2+81)32)2k′(x)=9(−2x2+81)(x2+81)52now, we can set it to 0 and then we can solve for xk′(x)=9(−2x2+81)(x2+81)52=0we will getk=922, x=−922we will get curvature maximum atR=922now, we can find y-valuey=9ln⁡(922)So, we will get(x,y)=(922,9ln⁡(922))limx→∞k′(x)=limx→∞9(−2x2+81)(x2+81)52we can see that degree of denominator is greaterSo, we getlimx→∞k′(x)=0

xleb123 · Accepted Answer

To find the point where the curve has maximum curvature for the equation y=9ln(x), we need to determine the value of x that corresponds to the maximum curvature. The curvature of a curve y=f(x) can be found using the formula:κ=|f″(x)|(1+f′(x)2)32where f″(x) represents the second derivative of f(x) and f′(x) represents the first derivative of f(x).For the given equation y=9ln(x), let&#039;s calculate the derivatives:f′(x)=ddx(9ln(x))=9xf″(x)=ddx(9x)=−9x2Substituting these values into the curvature formula, we get:κ=|−9x2|(1+(9x)2)32Simplifying further:κ=9x2(1+81x2)32To find the point of maximum curvature, we need to determine the value of x that maximizes κ. However, as x approaches infinity (x→∞), we can observe that the denominator of κ dominates the expression. The term (1+81x2)32 will approach 1 as x becomes very large, and κ will tend to 0.Hence, as x approaches infinity, the curvature (κ) of the curve y=9ln(x) tends to 0.

Jazz Frenia · Accepted Answer

The curvature of a function is given by the formula:κ=|y″|(1+(y′)2)32where y′ represents the first derivative of y with respect to x, and y″ represents the second derivative of y with respect to x.First, let&#039;s find the first derivative of y with respect to x:y′=ddx(9ln(x))=9·1x=9xNext, we find the second derivative of y with respect to x:y″=ddx(9x)=−9x2Substituting the values of y′ and y″ into the curvature formula, we get:κ=|−9x2|(1+(9x)2)32=9x2(1+81x2)32=9x21+81x2To find the maximum curvature, we need to determine the value of x that maximizes κ. To simplify the analysis, let&#039;s consider what happens to the curvature as x approaches infinity (x→∞).As x becomes infinitely large, the term 81x2 approaches zero. Thus, the curvature formula can be simplified as:κ≈9x21+0=9x2Therefore, as x approaches infinity, the curvature κ approaches zero. In other words, the curvature decreases and becomes negligible as x becomes very large.Hence, the curve has its maximum curvature at the starting point, and as x approaches infinity, the curvature decreases to zero.

Andre BalkonE · Accepted Answer

Step 1:The curvature (κ) of a curve defined by a function y=f(x) is given by the formula:κ=|f″(x)|[1+(f′(x))2]3/2where f′(x) represents the first derivative of f(x) with respect to x, and f″(x) represents the second derivative of f(x) with respect to x.First, let&#039;s find the first and second derivatives of the function y=9ln(x).The first derivative f′(x) is calculated as follows:f′(x)=ddx(9ln(x))=9xThe second derivative f″(x) is calculated by differentiating f′(x):f″(x)=ddx(9x)=−9x2Step 2:Now we have the necessary derivatives, we can substitute them into the formula for curvature:κ=|−9x2|[1+(9x)2]3/2Simplifying this expression, we get:κ=9x[1+(9x)2]3/2To find the point where the curve has maximum curvature, we need to find the value of x that maximizes κ. To do this, we can set the derivative of κ with respect to x equal to zero and solve for x:dκdx=0Differentiating κ with respect to x, we have:ddx(9x[1+(9x)2]3/2)=0By applying the chain rule and simplifying, we get:ddx(9x[1+(9x)2]3/2)=−9x2[1+(9x)2]3/2+27x3[1+(9x)2]5/2=0To solve this equation, we can multiply through by x3[1+(9x)2]5/2 to simplify the expression:−9[1+(9x)2]+27=0Expanding and rearranging the equation, we have:−9−81x−2+27=0Simplifying further, we get:81x−2−9=0Dividing through by 9, we have:9x−2−1=0Adding 1 to both sides, we get:9x−2=1Taking the reciprocal of both sides, we have:x2=19Taking the square root of both sides, we get:x=±13Step 3:Since x must be positive (as the natural logarithm is only defined for positive values), the solution is x=13.Therefore, the point where the curve y=9ln(x) has maximum curvature is when x=13.Now, let&#039;s analyze the behavior of the curvature as x approaches infinity (x→∞).As x becomes extremely large, we can observe the behavior of κ by taking the limit as x approaches infinity:limx→∞9x[1+(9x)2]3/2Simplifying this limit, we get:limx→∞9x[1+81x2]3/2As x tends to infinity, the term 81x2 approaches zero, and we can simplify the expression further:limx→∞9xTaking the limit of this expression as x approaches infinity, we have:limx→∞9x=0Therefore, as x approaches infinity, the curvature κ approaches zero.In summary, the curve y=9ln(x) has maximum curvature at x=13, and the curvature approaches zero as x tends to infinity.

At what point does the curve have maximum curvature? y= 9 \ln (x) What happens to the curvature as x \rightarrow \infty ?

Answered question

Answer & Explanation

New Questions in Calculus 1