Let C represent the parabolic cylinder's curve intersection&nbsp;x2=2y&nbsp;and the surface&nbsp;3z=xy. Calculate C's exact length from the starting point to the point&nbsp;(6,18,36).

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Let C represent the parabolic cylinder&#039;s curve intersection&amp;nbsp;x2=2y&amp;nbsp;and the surface&amp;nbsp;3z=xy. Calculate C&#039;s exact length from the starting point to the point&amp;nbsp;(6,18,36).

BleabyinfibiaG · Accepted Answer

In order to solve the first given equation for y in terms of x, let&#039;s first locate the curve of intersection, we getx2=2y/t, change the first equation into parametric form by replacing x with t, that isx=t,y=12t2Using t, solve the second z-equation we getz=13(x⋅y)=13(t⋅12t2)=16t3When we enter the curve&#039;s x, y, and z coordinates into the vector equation r(t), we getr(t)=&amp;lt;t,12t2,16t3&amp;gt;Calculate the first derivative of the vector equation r(t) component-wise, that is,r′(t)=&amp;lt;1,t,12t2&amp;gt;Calculate the magnitude of&amp;nbsp;r′(t)&amp;nbsp;and simplify.|r′(t)|=14t4+t2+1=12t4+4t2+4=12(t2+2)2=12t2+1Solve for the range of t along the curve between the origin and the point (6,18,36)(0,0,0)→t=0(6,18,36)→t=6∴0≤t≤6Set up an integral for the arc length from 0 to 6C=∫0612t2+1&amp;nbsp;dt&amp;nbsp;Evaluate the integralC=[16t3+t)]06=42Result: The exact length of curve C from origin to point (6,18,36) is 42

Let C be the curve intersection of the parabolic cylinder x^2=2y and the s

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