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  • Tell me? - Limit , continuity and differentiability - JEE Main-7

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 \left ( in\: \: cm/sec\right ) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1m above the ground is : 


 

  • Option 1)

    25 

  • Option 2)

     \frac{25}{3} 

  • Option 3)

     \frac{25}{\sqrt{3}} 

  • Option 4)

     25\sqrt{3}

 

Answers (1)

x^{2}+y^{2}=4                                   \frac{\mathrm{d} y}{\mathrm{d} t}=-25

x\frac{\mathrm{d} x}{\mathrm{d} t}+y\frac{\mathrm{d} y}{\mathrm{d} t}=0

\sqrt{3}\frac{\mathrm{d} x}{\mathrm{d} t}\pm \times 25=0

\frac{\mathrm{d} x}{\mathrm{d} t}=\frac{25}{\sqrt{3}}\: \: \: cm/sec


Option 1)

25 

Option 2)

 \frac{25}{3} 

Option 3)

 \frac{25}{\sqrt{3}} 

Option 4)

 25\sqrt{3}

Posted by

Vakul

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