Question

1) If y = x3 + 2x and dx/ dt = 5, find dy/dt when x = 2. Give only the numerical answer. For example, if dy, dt = 3, type only 3.

2)The area A = πr2 of a circular puddle changes with the radius. At what rate does the area change with respect to the radius when r = 5ft?


5π ft2/ft

5 ft2/ft

10π ft2/ft

25π ft2/ft


3)A 25-ft ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.18 ft/sec, how fast, in ft/sec, is the top of the ladder sliding down the wall, at the instant when the bottom of the ladder is 20 ft from the wall? Answer with 2 decimal places. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).

Answer 1

a) 70

b) 10π ft²/ft

c) 0.24 ft/sec

Explanation:

1) y = x³ + 2x

`(dy)/(dt)=(d(x^3+2x))/(dt)`

or

`(dy)/(dt)` =
`3x^2(dx)/(dt)+2(dx)/(dt)`

at
`(dx)/(dt)=5` and x = 2

`(dy)/(dt)` =
`3(2)^2*(5)+2*(5)`

or

`(dy)/(dt)` = 60 + 10 = 70

2) A = πr² ft²

`(dA)/(dr)=(d(\pi r^2))/(dr)`

or

`(dA)/(dr)`= 2(πr)

at r = 5 ft

`(dA)/(dr)`= 2(π × 5) ft²/ft

or

`(dA)/(dr)`= 10π ft²/ft

3) From Pythagoras theorem

Base² + Perpendicular² = Hypotenuse²

Thus,

B² + P² = H² .............(1)

here, H = length of the ladder

P is the height of the wall

B is the distance from the wall at bottom

or

B² + P² = 25² ...........(1)

at B = 20 ft

20² + P² = 25²

or

P² = 625 - 400

or

P = √225

or

P = 15 ft

differentiating (1) with respect to time, we get

`2B(dB)/(dt)+2P(dP)/(dt)=0`

at B = 20 ft,
`(dB)/(dt) = 0.18` and P = 15 ft

⇒ 2(20)(0.18) +
`2(15)(dP)/(dt)` = 0

or

`30(dP)/(dt)` = - 7.2

or

`(dP)/(dt)` = - 0.24 ft/sec (Here negative sign depicts the ladder slides down)