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Application of DerivativesPowerPoint Presentation

Application of Derivatives

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Application of Derivatives. Suppose a particle is moving in a straight line

Application of Derivatives

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- Suppose a particle is moving in a straight line
- Take ‘O’ as origin in time‘t’ the position of the particle be at A where OA= s and in time t+∆t the position of the particle be at B where OB=s+∆s clearly the directed distance of the particle from ‘o’ is function of time‘t’
- s=f (t)
- s+∆s=f (t+∆t)
- Then, average velocity of the particle between the points A(s) and B(s+∆s) is .
- The (instantaneous) velocity of the particle at the point A(s) at time t is the limiting value of as B A,
- i.e., as ∆t0, ∆s0

- Hence, the velocity at time
- Similarly, if v is the velocity of the particle at time t and v+∆v is the velocity at time t+, then average acceleration of the particle between the points s and is
- The acceleration at the point s is the limiting value of the average acceleration of the particle between the points
- Hence,

Algorithm

- Solving any problem first, see what are given data.
- What quantity is to be found?
- Find the relation between the point no. (1) & (2).
- Differentiate and calculate for the final result.

Example1 An edge of a variable cube is increasing at the rate of 10cm/sec. How fast the volume of the cube is increasing when the edge is 5 cm long?

- SOLUTION: Let x be the length of the edge of the cube and V be its volume at any time t. Then
- Given,
- Relation in V and x (as stated in the Algorithm)
- Now, V=x3
- (diff. Both Sides w.r.t.‘t’)
- =>

Example2:A stone is dropped into quiet lake and waves move in a circle at a speed of 3.5 cm/sec. At the instant when the radius of the circular wave is 7.5 cm. How fast is the enclosed area increasing?

- SOLUTION:Let r be the radius and A be the area of the circular wave at any time t. Then
- Relation in A and r
- Now: A
- => =
- =>=2
- =>

Example 3: A particle moves along the curve, 6y=x3+2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate

- SOLUTION: Let the required point be P(x, y). Then it is given that rate of change of y coordinate=8(rate of change of x-coordinate)
- i.e. 6y = x3 + 2
- => 6
- => 6
- => 3x2=48
- =>x2=16
- =>
- Now, x = 4
- =>6y = 43 + 2 = 66 i.e. y = 11
- And x = -4
- => 6y = (-4)3 + 2 = -62
- i.e. y =
- So, the required points are (-4, -31/3) and (4, 11).

Example 4: A man 2 meters high walks at a uniform speed of 6 meters per minute away from a lamp post, 5 meters high. Find the rate at which the length of his shadow increases.

- SOLUTION: Let AB be the lamp-post. Let at any time t, the man CD be at a distance x meters from the lamp-post and y meters be the length of his shadow CE. Then
- ……….(i)
- Now we try to find relation in x and y
- Now, triangle ABE and CDE are similar, therefore
- Diff. w.r.t. t
- Thus, the shadow increases at the rate of 4 meters/minute.

Example5: A ladder 5m long is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall, at the rate of 2m/sec. How fast its height on the wall decreasing when the foot of the ladder is 4m away from the wall?

- SOLUTION: Let AB be the position of the ladder at any time t such that OA= x and OB= y. Then,
- Given,
- ----------------------------- (i)
- =4 in (i), we get
- y
- So, The rate if decreasing in the height of the ladder on the wall is

Exercise

- A balloon which always remains spherical is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is 15 cm.
- A man 2 meters high walks at a uniform speed of 6Km/h away from a lamp-post 6 meters high. Find the rate at which the length of his shadow increases.
- The surface area of a spherical bubble is increasing at the rate of 2cm2/sec. when the radius of the bubble is 6cm, at what rate is the volume of the bubble increasing?