a) A boat is pulled into a dock by a rope attached to the bow ("front") of the boat and passing through a pulley on the dock that is 1m higher than the bow of the boat. If the rope is pulled in at a rate of 1m/s, how fast is the boat approaching the dock when it is 8m from the dock?
b)A particle is moving along the curve y=√x. As the particle passes through the point (4,2), its x-coordinate increases at a rate of 3cm/s. How fast is the distance from the particle to the origin changing at this instant?
c)The price (in dollars) p and the quantity demanded q are related by the equation: p2+2q2=1100.
If R is revenue, dR/dt can be expressed by the following equation: dR/dt=A(dp/dt),
where A is a function of just q.
i)Find A.
ii)Find dR/dt when q=20 and dp/dt=4.
I like apples
a) To solve this problem, we can use related rates.
Let's denote the distance between the boat and the dock as x, and the height of the pulley as h. Now, we can set up a right triangle with x as the base and (x+h) as the hypotenuse.
From the Pythagorean theorem, we have:
(x + h)^2 = x^2 + 1^2
x^2 + 2hx + h^2 = x^2 + 1
2hx = 1 - h^2
Now, we can differentiate both sides of the equation with respect to time (t):
2h(dx/dt) + 2x(dh/dt) = 0
Since we are given that the rope is being pulled in at a rate of 1m/s (dh/dt = 1), we can substitute that into the equation:
2h(dx/dt) + 2x(1) = 0
2h(dx/dt) = -2x
(dx/dt) = -x/h
To find the speed at which the boat is approaching the dock when it is 8m away, we need to substitute x = 8 into the equation and find the corresponding value of h.
Substituting x = 8 into the original equation: 2hx = 1 - h^2
16h = 1 - h^2
h^2 + 16h - 1 = 0
Using the quadratic formula, we find two possibilities for h, but since h cannot be negative in this context, we discard the negative value: h = √17 - 8.
Now we can substitute the values of x = 8 and h = √17 - 8 into (dx/dt) = -x/h:
(dx/dt) = -(8)/[(√17 - 8)]
(dx/dt) = -8/(√17 - 8)
Therefore, the boat is approaching the dock at a speed of 8/(√17 - 8) m/s when it is 8m from the dock.
b) To solve this problem, we can use the distance formula and related rates.
The distance between the particle and the origin can be given by the formula:
distance = √((x - 0)^2 + (y - 0)^2)
Substituting y = √x and x = 4, we have:
distance = √((4 - 0)^2 + (√4 - 0)^2)
distance = √(4 + 2)
distance = √6
We are given that the x-coordinate, x = 4, is increasing at a rate of 3 cm/s (dx/dt = 3 cm/s). We want to find d(distance)/dt, the rate at which the distance from the particle to the origin is changing.
Using the chain rule, we can differentiate the distance formula with respect to time (t):
d(distance)/dt = (1/2) * [2(x - 0)(dx/dt) + 2(y - 0)(dy/dt)]
Since we are given that dy/dt (the rate at which y is changing) is not provided, we can solve for it using the equation y = √x:
y = √x
dy/dt = (1/2) * (1/√x) * dx/dt
dy/dt = (1/2√x) * dx/dt
Substituting this into the equation for d(distance)/dt:
d(distance)/dt = (1/2) * [2(x - 0)(dx/dt) + 2(y - 0)((1/2√x) * dx/dt)]
d(distance)/dt = (1/2) * [2(x - 0)(dx/dt) + (1/√x) * dx/dt]
d(distance)/dt = (dx/dt) * (x + (1/2√x))
Substituting x = 4 and dx/dt = 3 into the equation:
d(distance)/dt = 3 * (4 + (1/2√4))
d(distance)/dt = 3 * (4 + 1)
d(distance)/dt = 3 * 5
d(distance)/dt = 15 cm/s
Therefore, the distance from the particle to the origin is changing at a rate of 15 cm/s when the particle passes through the point (4, 2).
c) i) To find A, we need to find the derivative of R with respect to q, while holding p constant.
Given the equation p^2 + 2q^2 = 1100, we differentiate both sides with respect to q:
2p(dp/dq) + 4q = 0
Solving for dp/dq, we have:
dp/dq = -2q/(2p) = -q/p
Therefore, A = -q/p.
ii) To find dR/dt when q = 20 and dp/dt = 4, we substitute these values into the equation dR/dt = A(dp/dt):
dR/dt = A(dp/dt)
dR/dt = (-q/p) * (dp/dt)
dR/dt = (-20/80) * 4
dR/dt = -1
Therefore, when q = 20 and dp/dt = 4, dR/dt is equal to -1.
a) To find how fast the boat is approaching the dock, we can use the concept of related rates. Let's assume the distance between the bow of the boat and the dock is x. Since the rope passes through a pulley that is 1m higher than the bow of the boat, the vertical distance between the rope and the dock is x + 1.
We need to find dx/dt, the rate at which x is changing when the boat is 8m from the dock. To do this, we can use the Pythagorean theorem to relate the variables:
x^2 + (x + 1)^2 = 8^2
Simplifying this equation, we get:
2x^2 + 2x - 63 = 0
Now, differentiate both sides of this equation with respect to time (t) using implicit differentiation:
4x(dx/dt) + 2(dx/dt) - 0 = 0
Solving for dx/dt:
6(dx/dt) = -2x
(dx/dt) = -x/3
Substituting x = 8 into the equation:
(dx/dt) = -(8)/3 = -8/3
Therefore, the boat is approaching the dock at a rate of 8/3 m/s.
b) To find how fast the distance from the particle to the origin is changing, we can use the concept of related rates. The equation representing the relationship between x and y is y = √x.
Let's assume the distance from the particle to the origin is r and the x-coordinate of the particle is x. We need to find dr/dt, the rate at which r is changing when x is increasing at a rate of 3 cm/s. To do this, we can use the Pythagorean theorem to relate the variables:
r^2 = x^2 + y^2
Substituting y = √x, we get:
r^2 = x^2 + (√x)^2 = x^2 + x
Differentiating both sides of this equation with respect to time (t) using implicit differentiation:
2r(dr/dt) = 2x(dx/dt) + dx/dt
Simplifying this equation:
dr/dt = (x + 1)dx/dt / r
Substituting x = 4, dx/dt = 3 cm/s, and r = 2 (since the particle is at (4,2)):
dr/dt = (4 + 1) * 3 / 2 = 15 / 2 = 7.5 cm/s
Therefore, the distance from the particle to the origin is changing at a rate of 7.5 cm/s.
c) i) To find A, we need to differentiate the equation representing the relationship between the price (p) and the quantity demanded (q):
p^2 + 2q^2 = 1100
Differentiating both sides of this equation with respect to time (t) using implicit differentiation:
2p(dp/dt) + 4q(dq/dt) = 0
Simplifying this equation:
4q(dq/dt) = -2p(dp/dt)
Dividing both sides of the equation by 4q:
(dq/dt) = -p(dp/dt) / (2q)
Since A is a function of just q, we can write A as:
A = -p / (2q)
ii) To find dR/dt when q = 20 and dp/dt = 4, we can use the equation:
dR/dt = A(dp/dt)
Substituting A = -p / (2q), q = 20, dp/dt = 4:
dR/dt = (-p / (2q)) * dp/dt = (-p / (2*20)) * 4 = -p/10
Therefore, when q = 20 and dp/dt = 4, dR/dt = -p/10.
If the boat is x meters from the dock, then the length of rope is given by
r^2 = x^2 + 1
2r r' = 2x x' = 2(8)(-1)
rr' = -8
So, r(8) = √65
r'√65 = -8
r' = 0.992 m/s
Makes sense; since the boat is far away as compared to the height, the rope length is essentially the distance away, so dr/dt is about the same as dx/dt.
The closer the boat gets, the slower the rope length decreases; when the boat is very near the dock, the rope length essentially stays at 1 meter.