All categories

3 years ago

Find the linear approximation of the function f(x, y, z) =

Mathematics

1 answer:

ser-zykov [4K]3 years ago

3 0

Answer:

See below in bold.

Step-by-step explanation:

f(x, y, z) = x^2 + y^2 + z^2 at (6, 6, 7)

Taking partial derivatives

f'x = 2x + y^2 + z^2 --> f'x(6,6,7) = 12+36+49= 97

f'y = x^2 + 2y + z^2---> f'y(6,6,7) = 36 + 12 + 49 = 97

f'z = x^2 + y^2 + 2z ----> f'z(6,6,7) = 36 + 36 + 14 = 86.

f(6,6,7) = 36 + 36 + 49 = 121

So our linear approximation is:

L(x,y,z) = 121 + 97(x - 6) + 97(y - 6) + 86(z - 7)

Substituting the given values

f(6.012, 5.972,6.982)

= L(6.012, 5.972, 6.982) ≈ 121 + 97(6.012 - 6) + 97(5.972 - 6) + 86(6.982 - 7)

= 117.9.

You might be interested in

What is -7.6 increased by 9?

Solnce55 [7]

Answer:

-7.6+9=1.4

Step-by-step explanation:

increased by is addition

6 0

3 years ago

P(5,3); y=4x<br><br>if you help me I will be thankful forever pls

horrorfan [7]

Y= -1/4x + 4.25, perdendicular then it must be the opposite of 4x which is -1/4. Then you multiply it with 5 and then add or subtract a number so you can find 3. So -1/4× 5 + 4.25= 3. Here you gooo!

6 0

3 years ago

7th grade Girl!

nika2105 [10]

Answer:

A, P, M, S, W, Q

4 0

3 years ago

On a science test, 2 points are deducted from a total of 100 points for each question answered incorrectly. Trevor scored an 84

ale4655 [162]

He answered 8 questions incorrectly. 2*8=16 100-16=84

6 0

3 years ago

Read 2 more answers

A cube-shaped aquarium has edges that are 3 feet long and is filled with water that has a density of 62 lbs/ft3 Should the aquar

Sunny_sXe [5.5K]

Given:

There are given the edge of the cube-shaped aquarium has 3 feet.

Explanation:

To find the value, we need to use the volume of the cube formula:

So,

From the formula of volume:

$V=a^3$

Where

a represents the value of edge.

So,

Put the value of edge into the above formula:

Then,

$\begin{gathered} V=a^3 \\ V=3^3 \\ V=27cube\text{ feet} \end{gathered}$

Now,

The water has a density of 62 pounds per cube foot.

According to the question:

If the weight of 1 cubic foot of water is 62 pounds, then the weight of 27 cube feet water is:

$62\times27=1674pound$

Final answer:

Hence, the water weight of the full aquarium is 1674 pounds, and the table only susupports00 pounds. So the table cannot hold the aquarium.

And,

No, the density of water would not change.

6 0

1 year ago