anyone good in calculus? here is a problem regarding relative rates please help i have a test tomorrow?

Question

a ladder 6.5 m long leans against a vertical wall. if the foot of the ladder slips outward at a rate of 1m/s , find the rate at which the top of the ladder is descending at the instant the foot is 2.5 m from the wall.
I have the answer...I need the solution though.....please help me , the answer is - 0.78m/s
thanks in advance to all who try..

yea i tried using d calculator comon am not that stupid...i knw how to do it just that I am not getting the right answer god knows why!!

hey hrhbg am getting the exact same answer....thanks for confirming!

Answer 1

By the Pythag. Theorem, the ladder is 6m high on the wall.

let x = distance from the wall, y = height on the wall, s = ladder
x=2.5, y=6, s=6.5 (distances) x'=1m/s, y'=??, s'=0 (rates of change)

x^2+y^2=s^2

implicit differentiation, 2xx'+2yy'=2ss'

so, 2(2.5)(1)+2(6)y'=0
5+12y'=0
12y'=-5
y'=-.417m/s

That's the answer I'm getting.

Answer 2

I agree with "hrhbg." I arrive at the same answer. I believe your book may have an error.

This situation needs to employ the Pythagorean Theorem, since the ladder forms a right triangle with the wall. Here's what you need to do first. Differentiate the equation (6.5)² = x² + y² implicitly:

(6.5)² = x² + y²
0 = 2 x x' + 2 y y'
-2 x x' = 2 y y'
-2 x x' / 2 y = y'

We are given x' = 1.5 m/sec and x = 2.5 m. From the first equation, we can calculate what y is given x = 2.5 m. Then all we have to do is plug those values into the last equation to solve for y'.

y = √[(6.5)² - (2.5)²]
y = √(42.25 - 6.25)
y = √36
y = 6.0 m.

Now we have everything we need to calculate y'.

y' = - 2 x x' / 2 y
y' = - 2 (2.5)(1.0) / 2 (6.0)
y' = - 5 /12
y' = - 0.417 m/s or - 0.42 m/s rounded to the nearest tenth of a meter per second.

Answer 3

The concentric around ripples grows in radius at sixteen cm/s. a million - The radius is sixteen*time = sixteen cm/seg * t 2 - The circule section is two*pi*radius^2 = 2pi r^2 We placed a million in 2 and we've 2pi * (sixteen t)^2 = 2 pi * 256 * t^2 the fee of section advance is the derivate of section in time dA/dt= 2pi *256 * 2 * t = 1024 * pi * t that's the fee for each and all of the posible cases. interior the specific time in wich the radius is 4 cm 0.25 seg (4cm / 16cm/seg) the fee is 1024 * pi *.25 = 256 * pi cm2/seg

Answer 4

I'm sorry...

No high school student needs to be learning that unless you will have a career in it.

we need to change our system

Answer 5

theres a thing called a calculator, waaaaaaaaaaaay easier, dude!!!!!

Answer 6

Is Anyone actually good in calculus???