need full working out for these multivariable calculus questions A space curve i

Question

need full working out for these multivariable calculus questions

A space curve is defined by C: r (u) = u^3 i + (5u^2 + 3) _j + 4uk. Find the Cartesian form of the equation for the plane that is perpendicular to the space curve C at the point where u = 4 Your answer should be an equation, expressed in terms of the Cartesian variables x, y and z using the correct syntax. For example: 3*x - 2*y + 5*z = 2, or, 2* (x - 1) + 4* (y - 2) + z - 1 = 0, or 3*x + 6*z = 12 - y, or y - x + 35* (z - 256) = 20 Do not use decimal approximations all numbers should be entered as exact expressions, for example 5/2

Explanation / Answer

C is given by r(u) = (u3, 5u2+3,4u)

Grad r = (3u^2, 10u, 4)

Grad r at u = 4 is

(48, 40,4)

Hence the plane perpendicular to the curve has normal with direction ratios (48,40,4) ~(12,10,1)

Also the plane passes through r(4) = (64, 83, 16)

Hence cartesian form of the plane that is perpendicular to the space curve C at u =4 is

12(x-64)+10(y-83)+1(z-16) =0

Simplify to get

12x+10y+z-768-830-16=0

Or

12x+10y+z - 1614=0

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