help with calculus rate problems?

Question

CAN SOMEONE PLEASE SHOW ME HOW TO DO THESE PROBLEMS... I HAVE THE ANSWER BUT HAVE NO IDEA HOW TO SOLVE THEM....

THANKS :)






17. A 13-foot ladder is leaning against a vertical wall. If the
bottom of the ladder is being pulled away from the wall
at the rate of 2 feet per second, how fast is the area of
the triangle formed by the wall, the ground, and the ladder
changing at the instant the bottom of the ladder is 12 feet
from the wall?



ANSWER: -119/5 feet squared per second




19. A tank contains 1000 cubic feet of natural gas at a pressure of 5 pounds per square inch. Find the rate of change
of the volume if the pressure decreases at a rate of 0.05
pounds per square inch per hour. (Assume Boyle’s law:
pressure × volume = constant.)

ANSWER: 10 feet cubed per hour

Answer 1

1.) Let, base be X
height of the ladder be Y
lets find Y when X is 12,
13^2 = X^2 + Y^2
Y^2= 169-144
Y = 5 when X = 12

X is increasing at the rate of 2 ft/sec (X' = 2 ft/sec)
now we need to know the rate of change of Y, when X is changing at 2 ft/sec

take the pythagoras eq again,
X^2 + Y^2 = 13^2
now lets take the derivative,
2X(X') + 2Y(Y') = 0
Substitute X= 12, Y = 5 and X' = 2 . we need Y'.
2(12)(2) + 2(5)(y') = 0
Y' = -48/10
Y'= -4.8
this means that Y is decreasing at the rate of 4.8 ft/sec

Area = 1/2 XY
we nee to find A'( rate of change of area)
A' = 1/2 (X'Y + Y'X)
A' = 1/2 (2*5 + (-4.8)(12))
A' = -23.8 ft/sec
or
A' = -119/5 ft/sec

2.) Volume is 1000 cu.ft when pressure is 5 pounds/sq.inch

P' = -0.05 pounds per sq.inch per hour
V' = ?
PV = K ,where k is a constant
take the derivative,
P'V + PV' = 0
(-0.05)(1000) + 5(V') = 0
V' = 50/5
V' = 10 cubic.ft /hour

Answer 2

Let´s see the first
The hight of the ladder is h= sqrt(13^2-x^2) where x is the distance of the bottom to the wall.
The area of the triangle is
A=1/2 x *sqrt(169-x^2)
dA/dt= dA/dx *dx/dt
=1/2 (sqrt(169-x^2)-x ^2* 1/sqrt(169-x^2))* dx/dt)

x=12 and dx/dt = 2
dA/dt =-119/5 feet squared/second
2) p v = K

p dv + vdp = 0 so p*dv/dt+v*dp/dt=0
5*dv/dt-1,000*0.05 = 0 (the minus sign arises as the pressure decreases)
so dv/dt = 10 cubic feet /sec

Answer 3

let the lower end of the ladder be A,the upper end be B,the point of intersection of the wall and the ground beC AC=X ,BC=Y
x^2+y^2+13^2 derivative both sides respect to time
2x dx/dt+2y dy/dt=0 *2and when x=12,AB=13 then y=5
12(2)+5(dy/dt)=0 then dy/dt=-24/5feet/sec.
let the area of triangleABCbe Z
Z=1/2(x)(y) derivative both sideswith respect to the time
dz/dt =1/2 (dx/d t )(y)+1/2(x)(dy/dt)
=(1/2)*(2)*(5)+(1/2)*(12)*(-24/5)=-23.8feet^2/sec.

Answer 4

[a] floor section = circumference * height = 2? r * h height = 10cm + 0.1cm * t h = 10 + 0.1t A = 10? * (10 + 0.1t) = 100? + ? t dA/dt = fee of substitute of section with relation to time dA/dt = (100? + ? t)' = ? cm^2/sec [b] volume = part of base * height = ? r^2 * (10 + 0.1t) V = 25? (10 + 0.1t) V = 250? + 2.5? t dV/dt = 2.5? cm^3/sec desire this enables!