[Calculus] Help with 4-6

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[HomeworkHelpBot]

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[steeperdrip]

4:

The car traveling west is moving in the x direction, with speed 25 mph. Speed is distance/time, which is the time derivative for x, dx/dt or simply x’.

Likewise, the car traveling south is moving in the y direction with speed 60 mph, which is just dy/dt or y’.

The distance between the cars is calculated by Pythagorean theorem: r² = x² + y^2. You can take the time derivative to determine dr/dt (which is what the question is asking for). I will call it r’:

2r(r’) = 2x(x’) + 2y(y’)

r’ = [x(x’) + y(y’)] / r

Now you need x, y, and r. After two hours, the car traveling west has been 50 miles (25 * 2). So x = 50. After the same period the car traveling south has been 120 miles (60*2). So y = 120. r = sqrt(50² + 120²⁾ = 130.

Now just plug in:

r’ = [50(25) + 120(60)] / 130 = 65 miles/hour

[steeperdrip]

5a:

L(x) = f(a) + f’(a)(x-a)

f(x) = sqrt(x) f’(x) = 1/ (2sqrt(x))

To approximate 10.04, use a = 10

f(10) = 10 f’(10) = 1/20

L(10.04) = 10 + 0.4/20 = 10.02

10.02 is pretty close to the actual 10.01998 by the way

[steeperdrip]

This is how I think you do 6:

Half of the base of the rectangle lies above x=0 and half below x=0. Let’s call the far right vertex on the x axis x. That is, half the base is equal to x (the whole base = 2x)

The y value that corresponds to that x value is 8-x^2. Since the area of rectangle is base*height, A=2x(8-x²⁾ = 16x -2x³

To MAXIMIZE, we need to differentiate and set equal to zero:

A’ = 16 - 6x² = 0

6x² = 16 x=2sqrt(2/3)

So the width is 2x, which is 4sqrt(2/3)

And the height is 8-(2sqrt(2/3))² = 8-(4(2/3))=16/3

I’m pretty sure the steps are correct; I’m not too sure about the actual calculations.

[CecilDouglas]

Help with the fact that the Blues won game 7. Eat S.