Question

(a) Find an equation of the tangent plane to the surface at the given point. xyz = 24, (1, 3, 8) (b) Find a set of symmetric equations for the normal line to the surface at the given point. $\frac{x+1}{24} = \frac{y+3}{3} = \frac{z+8}{8}$ $\frac{x}{24} = \frac{y}{3} = \frac{z}{8}$ $\frac{x-1}{24} = \frac{y-3}{8} = \frac{z-8}{3}$ $x-1 = y-3 = z-8$ $\frac{x+1}{24} = \frac{y+3}{8} = \frac{z+8}{3}$

          (a) Find an equation of the tangent plane to the surface at the given point.
xyz = 24, (1, 3, 8)
(b) Find a set of symmetric equations for the normal line to the surface at the given point.
$\frac{x+1}{24} = \frac{y+3}{3} = \frac{z+8}{8}$
$\frac{x}{24} = \frac{y}{3} = \frac{z}{8}$
$\frac{x-1}{24} = \frac{y-3}{8} = \frac{z-8}{3}$
$x-1 = y-3 = z-8$
$\frac{x+1}{24} = \frac{y+3}{8} = \frac{z+8}{3}$

(a) Find an equation of the tangent plane to the surface at the given point.
xyz = 24, (1, 3, 8)
(b) Find a set of symmetric equations for the normal line to the surface at the given point.
(x+1)/(24) = (y+3)/(3) = (z+8)/(8)
(x)/(24) = (y)/(3) = (z)/(8)
(x-1)/(24) = (y-3)/(8) = (z-8)/(3)
x-1 = y-3 = z-8
(x+1)/(24) = (y+3)/(8) = (z+8)/(3)

Added by Santiago K.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

08/30/2023

Step 1

Step 1: To find the equation of the tangent plane to the surface at the given point (1,3,8), we first need to find the gradient of the surface at that point. Show more…

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(a) Find an equation of the tangent plane to the surface at the given point. xyz=24,(1,3,8) (b) Find a set of symmetric equations for the normal line to the surface at the given point. (x+1)/(24)=(y+3)/(3)=(z+8)/(8) (x)/(24)=(y)/(3)=(z)/(8) (x-1)/(24)=(y-3)/(8)=(z-8)/(3) x-1=y-3=z-8 (x+1)/(24)=(y+3)/(8)=(z+8)/(3) a Find an equation of the tangent plane to the surface at the given point xyz=24,1,3,8 (b)Find a set of symmetric equations for the normal line to the surface at the given peint. x+1y+3z+8 24 3 8 Z 24 3 8 X1 Z-8 24 3 x1=y3 3=z-8 x+1 y+3 z+8 24 8 3