Find the curvature K of the curve at the point P. r(t)=ti+t2j+t34k, P(2,4,2)

Given r(t)=ti+t2j+t34k an point P(2,4,2)⇒x(t)=t, y(t)=t2 and z(t)=t34At P(2,4,2)x(t)=2⇒t=2Now,r′(t)=i+2tj+34t2kAt t=2r′(2)=i+2(2)j+34(2)2k=i+4j+3k⇒||r′(2)||=12+42+32=1+16+9=26r″(t)=2j+34(2t)k=2j+32tkAt t=2, r″(2)=2j+32(2)k=2j+3kr′(2)×r″(2)=(i+4j+3k)×(2j+3k)=6i−3j+2k⇒||r′(2)×r″(2)||=62+(−3)2+22=36+9+4=49=7Curvature of a curve at a given point is given byK=∣∣r′(t)×r″(t)||r′(t)||3⇒ at t=2, K=∣∣r′(t)×r″(t)||r′(t)||3=7(26)3=72626

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High School
Algebra
Pre-Algebra
Negative numbers and coordinate plane

Clifton Sanchez

Answered question

2021-11-15

Find the curvature K of the curve at the point P.

r (t) = t i + t^{2} j + \frac{t^{3}}{4} k, P (2, 4, 2)

Answer & Explanation

navratitehf

Beginner2021-11-16Added 20 answers

Given

r (t) = t i + t^{2} j + \frac{t^{3}}{4} k

an point P(2,4,2)

\Rightarrow x (t) = t, y (t) = t^{2}

and

z (t) = \frac{t^{3}}{4}

At P(2,4,2)

x (t) = 2 \Rightarrow t = 2

Now,

r^{'} (t) = i + 2 t j + \frac{3}{4} t^{2} k

At t=2

r^{'} (2) = i + 2 (2) j + \frac{3}{4} {(2)}^{2} k = i + 4 j + 3 k

\Rightarrow | | r^{'} (2) | | = \sqrt{1^{2} + 4^{2} + 3^{2}} = \sqrt{1 + 16 + 9} = \sqrt{26}

r^{″} (t) = 2 j + \frac{3}{4} (2 t) k = 2 j + \frac{3}{2} t k

t = 2, r^{″} (2) = 2 j + \frac{3}{2} (2) k = 2 j + 3 k

r^{'} (2) \times r^{″} (2) = (i + 4 j + 3 k) \times (2 j + 3 k) = 6 i - 3 j + 2 k

\Rightarrow | | r^{'} (2) \times r^{″} (2) | | = \sqrt{6^{2} + {(- 3)}^{2} + 2^{2}} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7

Curvature of a curve at a given point is given by

K = \frac{∣ ∣ r^{'} (t) \times r^{″} (t)}{{| | r^{'} (t) | |}^{3}}

\Rightarrow

t = 2, K = \frac{∣ ∣ r^{'} (t) \times r^{″} (t)}{{| | r^{'} (t) | |}^{3}} = \frac{7}{{(\sqrt{26})}^{3}} = \frac{7}{26 \sqrt{26}}

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