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Sonja [21]
2 years ago
12

Find an equation of the tangent plane to the surface z=ln(x-6y) at the point(7,1,0)

Mathematics
1 answer:
expeople1 [14]2 years ago
8 0

The tangent plane equation of tangent plane is

z = x - 6y -1

This is further explained below.

What is the equation of the tangent plane to the surface z=ln(x-6y) at the point(7,1,0)?

Generally, Given surface is z = ln(x − 6y) ,

Given point is (x_0, y_0 , z_0) = (7, 1, 0)

partially differentiating with respect to x

== >zx = [1/(x - 6y)] (1) = 1/(x - 6y)

zx (7 , 1 , 0) = 1/(7 - 6(1)) = 1

Partially differentiating with respect to y

zy = [1/(x - 6y)] (-6)

zy = -6/(x - 6y)

zy(7 , 1 , 0) = -6/(7 - 6(1)) = -6

Equation of tangent plane is z - z_0 = zx(x -x_0) + zy(y - y_0)

z - 0 = 1(x - 7) + (-6)(y - 1)

z = x - 7 -6y + 6

z = x - 6y -1

In conclusion, equation of tangent plane is z = x - 6y -1

Read more about the tangent plane

brainly.com/question/10542585

#SPJ4

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algol [13]

We need to differentiate this with respect to x to see if we can find an expression for the derivative of y at various points.  That will be the slope of the tangent to the curve.  Then we want to see where that derivative might be infinite -- i.e., where the tangent is vertical.

 

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(I used parentheses to show the differentiation of each term in the original equation.)

 

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So we have two x values where the tangent might be vertical.  Let's plug them into the equation and see what the y values are.  First x = 2...

 

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So at the point (2, -1) the tangent is vertical.

 

Now try x = -2...

 

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