1)A ladder is slipping down a vertical wall. If the ladder is 20ft long and the top of it is slipping at the constant rate of 2ft/sec, how fast is the bottom of the ladder moving along the ground when the bottom is 16ft from the wall?
**I know the answer is 1.5 ft/s (There's an answer key) But i don't know how to get to it! Can someone please explain!
2) The radius of a right circular cylinder is increasing at the rate of 6in/s while the height is decreasing at the rate of 3 in/s. At what rate is the volume of the cylinder changing when the radius is 5in and the height is 11in?
***Once again I know the answer just not how to get to it! The answer is 585pi in^3/s
**I know the answer is 1.5 ft/s (There's an answer key) But i don't know how to get to it! Can someone please explain!
2) The radius of a right circular cylinder is increasing at the rate of 6in/s while the height is decreasing at the rate of 3 in/s. At what rate is the volume of the cylinder changing when the radius is 5in and the height is 11in?
***Once again I know the answer just not how to get to it! The answer is 585pi in^3/s
-
Hello! it would have been better if I were able to draw so that you can easily understand.
1.) let z = the length of the ladder (this is constant)
x = the distance of the bottom of the ladder from the wall
y = the distance f the tip of the ladder from the ground
we know that:
dy/dt = -2ft/sec (I've put the negative sign here because it is approaching the ground)
we are looking for:
dx/dt
use pythagorean theorem to solve this since a right triangle is depicted here.
z^2 = x^2 + y^2
use power rule. since we know that z is a constant, its derivative would be zero.
z = 20^2
z' = 0
0 = 2x (dx/dt) +2y(dy/dt)
plug in the values:
0 = 2(16ft)dx/dt + 2(12 ft)(-2ft/sec) use pythagorean theorem to get the value of y here
dx/dt = 1.5 ft/sec
2. we know that the volume of a right circular cylinder is πr^2h.
given:
dr/dt = 6in/sec
dh/dt = -3in/sec again, i've put negative here because it is decreasing.
r = 5in
h = 11 in
v = πr^2h
use product rule here:
dv/dt = (2πr *dr/dt)(h) + (πr^2) (dh/dt)
plug in your values
dv/dt = (2π(5in) *(6in/sec))(11in) + (π5in^2) (-3in/sec)
dv/dt = 585πin^3/sec
:D
hope that helps!
1.) let z = the length of the ladder (this is constant)
x = the distance of the bottom of the ladder from the wall
y = the distance f the tip of the ladder from the ground
we know that:
dy/dt = -2ft/sec (I've put the negative sign here because it is approaching the ground)
we are looking for:
dx/dt
use pythagorean theorem to solve this since a right triangle is depicted here.
z^2 = x^2 + y^2
use power rule. since we know that z is a constant, its derivative would be zero.
z = 20^2
z' = 0
0 = 2x (dx/dt) +2y(dy/dt)
plug in the values:
0 = 2(16ft)dx/dt + 2(12 ft)(-2ft/sec) use pythagorean theorem to get the value of y here
dx/dt = 1.5 ft/sec
2. we know that the volume of a right circular cylinder is πr^2h.
given:
dr/dt = 6in/sec
dh/dt = -3in/sec again, i've put negative here because it is decreasing.
r = 5in
h = 11 in
v = πr^2h
use product rule here:
dv/dt = (2πr *dr/dt)(h) + (πr^2) (dh/dt)
plug in your values
dv/dt = (2π(5in) *(6in/sec))(11in) + (π5in^2) (-3in/sec)
dv/dt = 585πin^3/sec
:D
hope that helps!
-
1) Draw a right Triangle in the first quadrant of the plane. The hypotenuse of the Triangle is 20ft. If the ladder is Sliding down, y is decreasing while x is increasing. The relationship between x and y is x^2 + y^2 = 400. Find the time derivative of this expression: 2xdx/dt + 2ydy/dt = 0. dx/dt = -(y/x)dy/dt. You want to find dx/dt at the given conditions of the problem; the only additional thing you must be aware of is that y is decreasing, so, dy/dt = - 2ft/s dx/dt = -(12/16)( -2) = 1.5 ft/s
2) The volume V = (pi)r^2(h). Find the time rate of change of the volume V: dV/dt = (pi)( r^2dh/dt + 2rhdr/dt )
Substitute in the given values: dV/dt = (pi)( 5^2(-3) + 2(5)(11)(6) ) = ( -75 + 660 )(pi) = 585pi ft^3/s.
In each problem the negative rates had to be considered.
2) The volume V = (pi)r^2(h). Find the time rate of change of the volume V: dV/dt = (pi)( r^2dh/dt + 2rhdr/dt )
Substitute in the given values: dV/dt = (pi)( 5^2(-3) + 2(5)(11)(6) ) = ( -75 + 660 )(pi) = 585pi ft^3/s.
In each problem the negative rates had to be considered.