Application Of Derivatives

First, let's find the slope of the curve at the point (2, 3). We can do this by taking the derivative of the curve equation with respect to x:
2yy' = 3px²
Substituting the coordinates of the point (2, 3), we get:
2(3)y' = 3p(2²)
6y' = 12p
y' = 2p
Since the line y = 4x - 5 is tangent to the curve, the slope of the line at the point (2, 3) must be equal to the slope of the curve at that point. Therefore, we have:
2p = 4
Simplifying, we find:
p = 2
Now, let's substitute the value of p into the curve equation to find q:
y² = px³ + q
(3)² = (2)(2³) + q
9 = 16 + q
q = -7
Therefore, the values of p and q that satisfy the given conditions are p = 2 and q = -7.
The correct option is (B) p = 2, q = -7.

The given curve is , which can be rewritten as:

1. Finding the slope of the normal:
The slope of the tangent line to the curve at a given point is the derivative . For the normal line, the slope is the negative reciprocal of the tangent slope. Therefore, the slope of the normal is:

2. Equation of the tangent line:
The equation of the tangent line at a point on the curve is given by the point-slope form: Simplifying this equation: This is the equation of the tangent line to the curve at the point .

3. Condition for normal line:
For the line to be a normal to the curve at the point , it must be perpendicular to the curve at that point. The slope of the tangent to the curve at the point is given by: Therefore, the slope of the line perpendicular to the curve (the normal) is:

4. Product of slopes of tangent and normal:
The tangent and normal lines are perpendicular, so the product of their slopes must be . Hence, the product of the slopes is: Solving for , we get:

5. Substituting into the equation of the tangent line:
Substituting and into the equation of the tangent line , we get the two tangent lines:

6. Conclusion:
The normal at has a positive slope, and the normal at has a negative slope. Therefore, the correct option is:

(B)

We are asked to determine the rate of change of the volume of a sphere, given that the surface area is increasing at a specific rate.

Step 1: The formulas for the surface area and volume of a sphere:
The surface area of a sphere with radius is given by: The volume of a sphere with radius is given by:

Step 2: Given information:
We are told that the surface area is increasing at a rate of 2 cm²/sec: We are also asked to find the rate of change of the volume when the radius is 6 cm.

Step 3: Relating the rate of change of surface area to the rate of change of radius:
Differentiate the surface area formula with respect to time : This relates the rate of change of the surface area to the rate of change of the radius ( ).

Step 4: Relating the rate of change of volume to the rate of change of radius:
Similarly, differentiate the volume formula with respect to time : This relates the rate of change of the volume to the rate of change of the radius ( ).

Step 5: Substituting the known values:
We are given that cm²/sec, and we need to find when the radius is 6 cm.

Step 6: Calculating the rate of change of volume:
Now, substitute into the volume rate of change formula: Simplifying:

Final Answer:
Therefore, the rate of change of the volume when the radius is 6 cm and is increasing at a rate of cm/sec is .

Given:

• r is the radius of the cylinder, and h is its height.
• The volume V of the cylinder is given by:
  .
• The rate of change of the radius, (radius is increasing).
• The rate of change of the height, (height is decreasing).
• The values to be substituted: r = 3 cm, h = 5 cm.

We need to find:
when and

Step 1: Differentiate the volume formula with respect to time :
The formula for the volume of the cylinder is: Differentiating both sides with respect to , we get:

Step 2: Apply the product rule:
Since the volume formula involves a product of two functions and , we apply the product rule for differentiation: Next, differentiate with respect to : Therefore, we get:

Step 3: Substitute the given values:
Now, substitute the given values: , , , and : Simplifying:

Final Answer:
Therefore, the rate of change of volume is