A stone is dropped into a quiet lake and waves move in the form of circles at a speed of 4cm/sec. At the instant, when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing?
Answers
Step-by-step explanation:
Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ ᴛʜᴇ ᴀʀᴇᴀ ᴏғ ᴀ ᴄɪʀᴄʟᴇ ᴡɪᴛʜ ʀᴀᴅɪᴜs “ʀ” ɪs ɢɪᴠᴇɴ ʙʏ A = πʀ2.
Hᴇɴᴄᴇ, ᴛʜᴇ ʀᴀᴛᴇ ᴏғ ᴄʜᴀɴɢᴇ ᴏғ ᴀʀᴇᴀ “A’ ᴡɪᴛʜ ʀᴇsᴘᴇᴄᴛ ᴛᴏ ᴛʜᴇ ᴛɪᴍᴇ “ᴛ” ɪs ɢɪᴠᴇɴ ʙʏ:
ᴅA/ᴅᴛ = (ᴅ/ᴅᴛ) πʀ2
Bʏ ᴜsɪɴɢ ᴛʜᴇ ᴄʜᴀɪɴ ʀᴜʟᴇ, ᴡᴇ ɢᴇᴛ:
(ᴅ/ᴅʀ)(πʀ2). (ᴅʀ/ᴅᴛ) = 2πʀ.(ᴅʀ/ᴅᴛ)
Iᴛ ɪs ɢɪᴠᴇɴ ᴛʜᴀᴛ, ᴅʀ/ᴅᴛ = 4ᴄᴍ/sᴇᴄ
Tʜᴇʀᴇғᴏʀᴇ, ᴡʜᴇɴ ʀ = 10ᴄᴍ,
ᴅA/ᴅᴛ = 2π. (10). (4)
ᴅA.ᴅᴛ = 80 π
Hᴇɴᴄᴇ, ᴡʜᴇɴ ʀ = 10 ᴄᴍ, ᴛʜᴇ ᴇɴᴄʟᴏsɪɴɢ ᴀʀᴇᴀ ɪs ɪɴᴄʀᴇᴀsɪɴɢ ᴀᴛ ᴀ ʀᴀᴛᴇ ᴏғ 80π ᴄᴍ2/sᴇᴄ.
Hᴏᴘᴇ ɪᴛ's ʜᴇʟᴘ ᴜʜ ❤️
Answer:
Given as a stone is dropped into a quiet lake and waves move in circles at a speed of 4 cm/sec.
As to find the instant when the radius of the circular wave is 10 cm,
how fast is the enclosed area increasing Suppose r be the radius of the circle and A be the area of the circle
When ever stone is dropped into the lake waves moves in circle at speed of 4cm/sec.
that is the radius of the circle increases at a rate of 4cm/sec dr/dt = 4cm/sec ...(i)
As we know that the area of the circle is πr2
Therefore,
when the radius of the circular wave is 10 cm,
the above equation becomes dA/dt = 2π x 10 x 4 dA/dt = 80 πcm2/sec
Thus, the enclosed area is increasing at the rate of 80 πcm2/sec.
