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2 years ago

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

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Which points on the curve of x^2 - xy - y^2 = 5 have vertical tangent lines?

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.

It's not written as a function, but it can still be differentiated using the chain rule:

x2 + xy + y2 = 3

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

2x + x dy/dx + y + 2y dy/dx = 0

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2x + y = dy/dx (-x -2y)

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We have the derivative of y, but it's defined partly in terms of y itself. That's OK. Let's go on...

So where would the slope be infinite? That would happen when x + 2y = 0, or y = -x/2

Let's plug that in for y in the original equation to find points where that's the case.

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x2 - x2/2 + x2/4 = 3

3x2 / 4 = 3

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x = ±2

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...

x2 + xy + y2 = 3

4 + 2y + y2 = 3

y2 + 2y + 1 = 0

(y + 1)2 = 0

y = -1

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

Now try x = -2...

x2 + xy + y2 = 3

4 - 2y + y2 = 3

y2 - 2y + 1 =0

(y - 1)2 = 0

y = 1

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

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3 years ago

PLEASE ANSWER ASAP !!!!!

VLD [36.1K]

k(6)=0

#pglubestiehere

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