Application Of Derivatives

Perimeter of sector =
(given)


Area of sector,



For maximum area,


$\therefore \frac{d^{2}A}{dr^{2}} = \frac{\pi}{180} \left(0-2\right) = \frac{-\pi}{90}

Given, curve ...(i)
Let the coordinate of point of contact

Now, equation of tangent at 'P'

Since, the tangent passing through the origin ie,

At ' from E (i)





So, point of contact is .
Now, slope of normal

Equation of normal at 'P'


...(ii)
The length of perpendicular drawn from the origin to the normal

The given curve is


We know that
Length of subnormal
Length of subtangent
But given condition is

Given that of the wire
Then, diameter + arc length


...(i)
Also, know that area of semicircle
...(ii)

Angle

for straight length of wire
...(iii)
From E (i)

...(iv)
Now,
For max or min area of enclosed by wire


Then, from E (iv)

We know that, is increasing (strictly) When





Hence,

Given,
Here,
Centre =
Radius




i.e., The distance between the points (2, 1) and


units
The least distance of the point (10, 7) from the circle = 10 - 5 = 5 units and the greatest distance of the point (10, 7) from the circle = 10 + 5 = 15 units.

Given,
Taking logarithm on both sides, we get
log
log a

On differentiating E (i) w.r.t. 'x', we get



Length of subtangent



Length of sub tangent at

Given,
Area of circular wave,

We have,



Now,

Let





For maxima,




Hence, the maximum value of is
i.e., at

We have, (i)
Put tangent crosses -axis]
On differentiating E (i) w.r.t. , we get



So, the slope of normal is .
Equation of the normal at is

We have,


Now, [say]
and


Now, [say]

Hence, both the curves cut each other at right angle.

We have,


Slope of tangent
Now, tangent makes an angle of with -axis
Slope of tangent
From Eqs. (ii) and (iii), we get


Putting, in E (i), we get


Required point is

Take

Step 1: Understanding the Concept:
This is an application of derivatives involving rates of change. The phrase "y-coordinate changes 9 times as fast as x-coordinate" translates mathematically to a specific relationship between their derivatives with respect to time ( and ).

Step 2: Key Formula or Approach:
1. Express the given rate condition mathematically: . 2. Differentiate the given curve equation with respect to time . 3. Substitute the rate condition into the differentiated equation to solve for . 4. Substitute the found values back into the original curve equation to find the corresponding coordinates.

Step 3: Detailed Explanation:
The equation of the drone's path is: Differentiate both sides with respect to time : Divide both sides by 3: We are given the condition that the y-coordinate changes 9 times as fast as the x-coordinate. This means: Substitute this condition into our differentiated equation: Assuming the drone is actually moving, , so we can divide both sides by : This gives two possible values for the x-coordinate: Now, find the corresponding y-coordinates by substituting these x values back into the original path equation . Case 1: If : So, one position is . Case 2: If : So, another position is . Looking at the given options, is not present, but is.

Step 4: Final Answer:
One of the precise positions is .

Step 1: Understanding the Concept:
We need to find the time at which a given function , representing the number of shares, reaches its maximum value. This is a classic optimization problem solvable using the first and second derivative tests.

Step 2: Key Formula or Approach:
1. Find the first derivative . 2. Set to find the critical points (potential maxima or minima). 3. Find the second derivative . 4. Evaluate at the critical points. A negative value indicates a local maximum.

Step 3: Detailed Explanation:
The function for the number of shares is: Calculate the first derivative with respect to : To find critical points, set the first derivative to zero: Factor the equation: This gives two critical points: and . Since time must be meaningful in the context of the problem (usually for something "getting viral" after posting), is our primary candidate. Let's verify it's a maximum. Calculate the second derivative: Evaluate the second derivative at the critical points: At : . This indicates a local minimum. At : . Since the second derivative is negative, the function attains a local maximum at this point. Therefore, the maximum number of shares occurs at hour.

Step 4: Final Answer:
The video gets the maximum number of shares at .