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AlekseyPX
3 years ago
15

Derivative of cos(2t) is?

Mathematics
1 answer:
miskamm [114]3 years ago
6 0


The derivative of the function cos(2t) can be determined by following the trigonometric rule of differentiation. In this case, the derivative of cos t is -sin t while that of 2t is t. The total derivative and the answer to this problem asking for the  d<span>erivative of cos(2t) </span><span>is -2 sin 2t.</span>
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Step-by-step explanation:

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Find the extremum of f(x,y) subject to the given constraint, and state whether it is a maximum or a minimum. f(x,y)=4x2 2y2; 3x
nevsk [136]

For given f(x, y) the extremum: (12, 24) which is the minimum.

For given question,

We have been given a function f(x) = 4x² + 2y² under the constraint 3x+3y= 108

We use the constraint to build the constraint function,

g(x, y) = 3x + 3y

We then take all the partial derivatives which will be needed for the Lagrange multiplier equations:

f_x=8x

f_y=4y

g_x=3

g_y=3

Setting up the Lagrange multiplier equations:

f_x=\lambda g_x

⇒ 8x = 3λ                                        .....................(1)

f_y=\lambda g_y

⇒ 4y = 3λ                                         ......................(2)

constraint: 3x + 3y = 108                .......................(3)

Taking (1) / (2), (assuming λ ≠ 0)

⇒ 8x/4y = 1

⇒ 2x = y

Substitute this value of y in equation (3),

⇒ 3x + 3y = 108

⇒ 3x + 3(2x) = 108

⇒ 3x + 6x = 108

⇒ 9x = 108

⇒ x = 12

⇒ y = 2 × 12

⇒ y = 24

So, the saddle point (critical point) is (12, 24)

Now we find the value of f(12, 24)

⇒ f(12, 24) = 4(12)² + 2(24)²

⇒ f(12, 24) = 576 + 1152

⇒ f(12, 24) = 1728                             ................(1)

Consider point (18,18)

At this point the value of function f(x, y) is,

⇒ f(18, 18) = 4(18)² + 2(18)²

⇒ f(18, 18) = 1296 + 648

⇒ f(18, 18) = 1944                            ..............(2)

From (1) and (2),

1728 < 1944

This means, given extremum (12, 24) is minimum.

Therefore, for given f(x, y) the extremum: (12, 24) which is the minimum.

Learn more about the extremum here:

brainly.com/question/17227640

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6 0
2 years ago
On a town map, each unit of the coordinate plane represents 1 mile. Three branches of a bank are located at A(−3, 1), B(3, 3), a
Andrei [34K]

Answer:

The minimum total distance the employee may have driven before getting stuck in traffic is 8.3 miles

Step-by-step explanation:

* Lets explain how to solve the problem

- The rule of the distance between to points (x_{1},y_{1})

  and (x_{2},y_{2}) i s d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}

- So at first we will find the distance between the three points

∵ A = (-3 , 1) , B = (3 , 3) , C = (3 , -1)

- By using the rule of distance

∴ AB=\sqrt{(3--3)^{2}+(3-1)^{2}}=\sqrt{36+4}=\sqrt{40}=6.325

∴ BC=\sqrt{(3-3)^{2}+(-1-3)^{2}}=\sqrt{0+16}=4

∴ AC=\sqrt{(3--3)^{2}+(-1-1)^{2}}=\sqrt{36+4}=\sqrt{40}=6.325

∵ Each unit coordinate plane represents 1 mile

∵ AB = 6.3 units

∴ The distance between branches A and B = 6.325 miles

∵ BC = 4 units

∴ The distance between branches B and C = 4 miles

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∴ The distance between branches A and C = 6.325 miles

- A bank employee drives from Branch A to Branch B and then

 drives halfway to Branch C before getting stuck in traffic

∵ The distance from branch A to branch B is 6.325 miles

∵ The distance from branch B to branch C is 4 miles

∵ The half way from branch B to branch C = 1/2 × 4 = 2 miles

∴ The distance the employee may have driven before getting

   stuck in traffic = 6.325 + 2 = 8.325 miles

∴ The minimum total distance the employee may have driven

   before getting stuck in traffic is 8.3 miles

8 0
3 years ago
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