Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        The top of ladder 6m long is resting against a vertical wall on a level pavement ,when the ladder begins to slide outwards .At the moment when the foot of ladder is 4m from the wall ,it is sliding away from the wall at the rate of 0.5m/sec. how fast is the top-sliding downwards at instanse? How far is the foot the wall when it and the top are moving at the same rate`
7 years ago

Prudhvi teja
83 Points
```										Dear rohit
x^2+y^2= l^2
x is the foot distance and y is the height at which top point of ladder touched wall
differentin you can get
2xdx/dt = -2ydy/dt
x=4
l=6
dx/dt = 0.5
so y = 4.472
dy/dt=0.447

In the same way for the second part dy/dt = dx/dt
so y=x
so y=x =3*2^0.5

Please         feel free to post as many doubts on our  discussion forum as you       can.we   will get you the answer and  detailed  solution very    quickly.All the best.

Now you can win exciting gifts by          answering the questions on Discussion Forum. So help discuss  any       query   on askiitians forum and become an Elite Expert League     askiitian

```
7 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Differential Calculus

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details