dtdy=−25cm/sec , dtdx=?
Now, x2+y2=22=4
differentiating w.r.t. t both side
2xdtdx+2ydtdy=0 x2+y2=4 when y=1 ⇒x2=3⇒x=3
⇒dtdx=−xydtdy
⇒dtdx(3,1)=−31×(−25)=325cm/sec
JEE Main 2019 — Mathematics Calculus
A 2m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25cm/sec , then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is:
Held on 12 Apr 2019 · Verified 6 Jul 2026.
25
253
325
325
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