45.1 Curvature Problem Set
CURVATURE | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: The radius of curvature at the point (1, −1) on the curve 𝑦 = 𝑥 ! − 3𝑥 + 1 is most close to: A. 9 B.
3
C.
2
D. 14
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PROBLEM 2: The radius of curvature at any point on the curve 𝑓 𝑥 = 𝑎𝑙𝑜𝑔(sec represented as:
A. (𝑎)(cos B. (𝑎)(cot C. (𝑎)(tan D. (𝑎)(sec
! ! ! ! ! ! ! !
) ) ) )
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! !
) is best
PROBLEM 3: The radius of curvature of the function 𝑓 𝑥 = 2𝑙𝑜𝑔𝑠𝑖𝑛
! !
at 𝑥 =
! !
A. −4 B. 3 C. 22 D. 4
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is most close to:
CURVATURE | SOLUTIONS SOLUTION 1: The topic of RADIUS OF CURVATURE can be referenced on page 28 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
The RADIUS OF CURVATURE “𝑅” at any point on a curve is defined as the absolute value of the reciprocal of the curvature “𝐾” at that same point. In formulaic terms:
𝑅=
1 𝐾
Where 𝐾 ≠ 0 Recall that K represents the CURVATURE and is generally written as:
𝐾=
𝑦" 1 + 𝑦!
! ! !
Taking this formula for CURVATURE and putting it in to this context of the RADIUS OF CURVATURE, we have:
𝑅=
1 + 𝑦! |𝑦"|
! ! !
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Where 𝑦" ≠ 0 In this problem, we are given a function: 𝑓(𝑥) = 𝑥 ! − 3𝑥 + 1 We are asked to determine the RADIUS OF CURVATURE of this function at the point (1, −1). To start us off, we first need to determine the derivative of 𝑓(𝑥) which is: 𝑓 ! 𝑥 = 2𝑥 − 3 Referring back to our RADIUS OF CURVATURE general formula, we see that this function will be squared, so carrying this calculation out, we get: [𝑓 ! 𝑥 ]! = (2𝑥 − 3)! = 4𝑥 ! − 12𝑥 + 9 We also need the Second Derivative, which is: 𝑓"(𝑥) = 2
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We now have all the data we need to start plugging in to our general formula for the RADIUS OF CURVATURE, which again reads:
𝑅=
1 + 𝑦! |𝑦"|
! ! !
Substitute our values in we get: [1 + 4𝑥 ! − 12𝑥 + 9]!/! 𝑅= 2 And simplifying a bit we have: [4𝑥 ! − 12𝑥 + 10]!/! 𝑅= 2 At this point, all we need to do to determine the RADIUS OF CURVATURE at the desired point “𝑥” is to plug in the value of 1, which gives us: [4(1)! − 12(1) + 10]!/! 𝑅= 2 We calculates out as: {2]!/! 𝑅= = 2 2
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The correct answer choice is C. 𝟐
SOLUTION 2: The topic of RADIUS OF CURVATURE can be referenced on page 28 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
The RADIUS OF CURVATURE “𝑅” at any point on a curve is defined as the absolute value of the reciprocal of the curvature “𝐾” at that same point. In formulaic terms:
𝑅=
1 𝐾
Where 𝐾 ≠ 0 Recall that K represents the CURVATURE and is generally written as:
𝐾=
𝑦" 1 + 𝑦!
! ! !
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Taking this formula for CURVATURE and putting it in to this context of the RADIUS OF CURVATURE, we have:
𝑅=
1 + 𝑦! |𝑦"|
! ! !
Where 𝑦" ≠ 0 In this problem, we are given a function: 𝑥 ) 𝑎
𝑓 𝑥 = 𝑎𝑙𝑜𝑔(sec
We are asked to determine a general statement of the RADIUS OF CURVATURE for this function, and not at a specific point. To start us off, we first need to determine the derivative of 𝑓(𝑥) which is: !
𝑓! 𝑥 = 𝑎 𝑥
!"#
! !
∙ sec
! !
tan
! !
!
!
∙ ! = tan (!)
We also need the Second Derivative, which is: !
!
𝑓"(𝑥) = 𝑠𝑒𝑐 ! (!) ∙ !
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We now have all the data we need to start plugging in to our general formula for the RADIUS OF CURVATURE, which again reads:
𝑅=
1 + 𝑦! |𝑦"|
! ! !
Substitute our values in we get: 𝑥 ! [1 + tan (𝑎) ]!/! 𝑅= 𝑥 1 𝑠𝑒𝑐 ! (𝑎) ∙ 𝑎 And simplifying a bit we have:
𝑅 = (𝑎)(sec
𝑥 ) 𝑎
The correct answer choice is D. (𝒂) 𝒔𝒆𝒄
𝒙 𝒂
SOLUTION 3: The topic of RADIUS OF CURVATURE can be referenced on page 28 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
The RADIUS OF CURVATURE “𝑅” at any point on a curve is defined as the absolute value of the reciprocal of the curvature “𝐾” at that same point.
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In formulaic terms:
𝑅=
1 𝐾
Where 𝐾 ≠ 0 Recall that K represents the CURVATURE and is generally written as: 𝑦"
𝐾= 1+
! ! ! ! 𝑦
Taking this formula for CURVATURE and putting it in to this context of the RADIUS OF CURVATURE, we have:
𝑅=
! ! ! ! 𝑦
1+ |𝑦"|
Where 𝑦" ≠ 0 In this problem, we are given a function:
𝑓 𝑥 = 2𝑙𝑜𝑔𝑠𝑖𝑛
𝑥 2
We are asked to determine the RADIUS OF CURVATURE of this function at the point !
𝑥 = !.
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To start us off, we first need to determine the derivative of 𝑓(𝑥) which is:
𝑓! 𝑥 = 2 ∙
1
𝑥 1 𝑥 ∙ cos ∙ = cot ( ) 𝑥 2 2 2 sin 2
We also need the Second Derivative, which is: 𝑥 1 𝑓"(𝑥) = −𝑐𝑠𝑐 ! ( ) ∙ 2 2 !
Evaluating each of the derivatives at we get: !
𝑓!
𝜋 𝜋 = cot = 3 3 6
And: 𝜋 1 𝜋 𝑓"( ) = − 𝑐𝑠𝑐 ! ( ) = −2 3 2 6 We now have all the data we need to start plugging in to our general formula for the RADIUS OF CURVATURE, which again reads:
𝑅=
1 + 𝑦! |𝑦"|
! ! !
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Substitute our values in we get:
𝑅=
[1 +
!
3 ]!/! −2
Which calculates out as: 𝑅 = −4 The correct answer choice is A. −𝟒
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PROBLEM 1: The radius of curvature at the point (1, −1) on the curve 𝑦 = 𝑥 ! − 3𝑥 + 1 is most close to: A. 9 B.
3
C.
2
D. 14
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PROBLEM 2: The radius of curvature at any point on the curve 𝑓 𝑥 = 𝑎𝑙𝑜𝑔(sec represented as:
A. (𝑎)(cos B. (𝑎)(cot C. (𝑎)(tan D. (𝑎)(sec
! ! ! ! ! ! ! !
) ) ) )
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! !
) is best
PROBLEM 3: The radius of curvature of the function 𝑓 𝑥 = 2𝑙𝑜𝑔𝑠𝑖𝑛
! !
at 𝑥 =
! !
A. −4 B. 3 C. 22 D. 4
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is most close to:
CURVATURE | SOLUTIONS SOLUTION 1: The topic of RADIUS OF CURVATURE can be referenced on page 28 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
The RADIUS OF CURVATURE “𝑅” at any point on a curve is defined as the absolute value of the reciprocal of the curvature “𝐾” at that same point. In formulaic terms:
𝑅=
1 𝐾
Where 𝐾 ≠ 0 Recall that K represents the CURVATURE and is generally written as:
𝐾=
𝑦" 1 + 𝑦!
! ! !
Taking this formula for CURVATURE and putting it in to this context of the RADIUS OF CURVATURE, we have:
𝑅=
1 + 𝑦! |𝑦"|
! ! !
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by Prepineer | Prepineer.com
Where 𝑦" ≠ 0 In this problem, we are given a function: 𝑓(𝑥) = 𝑥 ! − 3𝑥 + 1 We are asked to determine the RADIUS OF CURVATURE of this function at the point (1, −1). To start us off, we first need to determine the derivative of 𝑓(𝑥) which is: 𝑓 ! 𝑥 = 2𝑥 − 3 Referring back to our RADIUS OF CURVATURE general formula, we see that this function will be squared, so carrying this calculation out, we get: [𝑓 ! 𝑥 ]! = (2𝑥 − 3)! = 4𝑥 ! − 12𝑥 + 9 We also need the Second Derivative, which is: 𝑓"(𝑥) = 2
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We now have all the data we need to start plugging in to our general formula for the RADIUS OF CURVATURE, which again reads:
𝑅=
1 + 𝑦! |𝑦"|
! ! !
Substitute our values in we get: [1 + 4𝑥 ! − 12𝑥 + 9]!/! 𝑅= 2 And simplifying a bit we have: [4𝑥 ! − 12𝑥 + 10]!/! 𝑅= 2 At this point, all we need to do to determine the RADIUS OF CURVATURE at the desired point “𝑥” is to plug in the value of 1, which gives us: [4(1)! − 12(1) + 10]!/! 𝑅= 2 We calculates out as: {2]!/! 𝑅= = 2 2
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The correct answer choice is C. 𝟐
SOLUTION 2: The topic of RADIUS OF CURVATURE can be referenced on page 28 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
The RADIUS OF CURVATURE “𝑅” at any point on a curve is defined as the absolute value of the reciprocal of the curvature “𝐾” at that same point. In formulaic terms:
𝑅=
1 𝐾
Where 𝐾 ≠ 0 Recall that K represents the CURVATURE and is generally written as:
𝐾=
𝑦" 1 + 𝑦!
! ! !
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by Prepineer | Prepineer.com
Taking this formula for CURVATURE and putting it in to this context of the RADIUS OF CURVATURE, we have:
𝑅=
1 + 𝑦! |𝑦"|
! ! !
Where 𝑦" ≠ 0 In this problem, we are given a function: 𝑥 ) 𝑎
𝑓 𝑥 = 𝑎𝑙𝑜𝑔(sec
We are asked to determine a general statement of the RADIUS OF CURVATURE for this function, and not at a specific point. To start us off, we first need to determine the derivative of 𝑓(𝑥) which is: !
𝑓! 𝑥 = 𝑎 𝑥
!"#
! !
∙ sec
! !
tan
! !
!
!
∙ ! = tan (!)
We also need the Second Derivative, which is: !
!
𝑓"(𝑥) = 𝑠𝑒𝑐 ! (!) ∙ !
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We now have all the data we need to start plugging in to our general formula for the RADIUS OF CURVATURE, which again reads:
𝑅=
1 + 𝑦! |𝑦"|
! ! !
Substitute our values in we get: 𝑥 ! [1 + tan (𝑎) ]!/! 𝑅= 𝑥 1 𝑠𝑒𝑐 ! (𝑎) ∙ 𝑎 And simplifying a bit we have:
𝑅 = (𝑎)(sec
𝑥 ) 𝑎
The correct answer choice is D. (𝒂) 𝒔𝒆𝒄
𝒙 𝒂
SOLUTION 3: The topic of RADIUS OF CURVATURE can be referenced on page 28 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
The RADIUS OF CURVATURE “𝑅” at any point on a curve is defined as the absolute value of the reciprocal of the curvature “𝐾” at that same point.
Made with
by Prepineer | Prepineer.com
In formulaic terms:
𝑅=
1 𝐾
Where 𝐾 ≠ 0 Recall that K represents the CURVATURE and is generally written as: 𝑦"
𝐾= 1+
! ! ! ! 𝑦
Taking this formula for CURVATURE and putting it in to this context of the RADIUS OF CURVATURE, we have:
𝑅=
! ! ! ! 𝑦
1+ |𝑦"|
Where 𝑦" ≠ 0 In this problem, we are given a function:
𝑓 𝑥 = 2𝑙𝑜𝑔𝑠𝑖𝑛
𝑥 2
We are asked to determine the RADIUS OF CURVATURE of this function at the point !
𝑥 = !.
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To start us off, we first need to determine the derivative of 𝑓(𝑥) which is:
𝑓! 𝑥 = 2 ∙
1
𝑥 1 𝑥 ∙ cos ∙ = cot ( ) 𝑥 2 2 2 sin 2
We also need the Second Derivative, which is: 𝑥 1 𝑓"(𝑥) = −𝑐𝑠𝑐 ! ( ) ∙ 2 2 !
Evaluating each of the derivatives at we get: !
𝑓!
𝜋 𝜋 = cot = 3 3 6
And: 𝜋 1 𝜋 𝑓"( ) = − 𝑐𝑠𝑐 ! ( ) = −2 3 2 6 We now have all the data we need to start plugging in to our general formula for the RADIUS OF CURVATURE, which again reads:
𝑅=
1 + 𝑦! |𝑦"|
! ! !
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Substitute our values in we get:
𝑅=
[1 +
!
3 ]!/! −2
Which calculates out as: 𝑅 = −4 The correct answer choice is A. −𝟒
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