A particle moves along the curve y=lnx so that its abscissa is increasing at a rate of 2 units per second. At what rate is the particle moving away from the origin as it passes through the point (1,e)?
y = lnx
The distance z from the origin is thus
z^2 = x^2 + (lnx)^2
2z dz/dt = 2x dx/dt + 2lnx (1/x) dx/dt
z dz/dt = (x+lnx/x) dx/dt
at (1,e), z = √(1+e^2), so
dz/dt = (1+1/1)(2)/√(1+e^2) = 4/√(1+e^2)
Well, let's break this down. The particle is moving along the curve y=lnx, which means that the y-coordinate is given as the natural logarithm of the x-coordinate.
Since the abscissa (x-coordinate) is increasing at a rate of 2 units per second, we can say that dx/dt = 2.
We need to find the rate at which the particle is moving away from the origin, which means we need to find dy/dt when x=1 and y=e.
To find dy/dt, we need to use the chain rule. The chain rule states that if y = f(u) and u = g(x), then dy/dx = (dy/du) * (du/dx).
In this case, we have y = ln(x) and x=1. So, we have u = g(x) = x and y = f(u) = ln(u). Therefore, du/dx = 1 and dy/du = 1/u.
Using the chain rule, dy/dx = (dy/du) * (du/dx) = (1/u) * 1 = 1/u.
Now, we can substitute the values of u and y into the equation to find dy/dt. When x=1, u=1 and y=e.
So, dy/dt = (1/u) * (du/dt) = (1/1) * 2 = 2.
Therefore, the particle is moving away from the origin at a rate of 2 units per second as it passes through the point (1,e).
To find the rate at which the particle is moving away from the origin, we need to find the rate of change of the distance between the particle and the origin. We can do this by using the distance formula:
Distance = √((x - 0)^2 + (y - 0)^2)
Since the particle moves along the curve y = ln(x), we can substitute this into the distance formula:
Distance = √((x - 0)^2 + (ln(x) - 0)^2)
To find the rate at which the particle is moving away from the origin, we need to differentiate this expression with respect to time (t). Let's denote the distance between the particle and the origin as D:
D = √(x^2 + (ln(x))^2)
Differentiating both sides of this equation with respect to time:
dD/dt = d/dt √(x^2 + (ln(x))^2)
Now, let's find the values of x and y when the particle passes through the point (1, e). Substituting these values into the equation for the distance, we get:
D = √((1 - 0)^2 + (ln(1) - 0)^2) = √(1 + 0) = 1
Now, we can substitute the given rate at which the abscissa is increasing into the equation for dD/dt:
2 = d/dt √(x^2 + (ln(x))^2)
To find dD/dt, we need to differentiate the expression inside the square root:
d/dt (x^2 + (ln(x))^2)
= 2x(dx/dt) + 2(ln(x))(d(ln(x))/dt)
Since dx/dt is given as 2 and d(ln(x))/dt can be found using the chain rule:
d(ln(x))/dt = 1/x(dx/dt) = 1/x(2) = 2/x
Substituting these values into the equation for dD/dt:
2 = 2x(2) + 2(ln(x))(2/x)
2 = 4x + 4(ln(x))
2 - 4x = 4(ln(x))
To find the value of x, we can solve the equation using logarithmic properties:
ln(x^2) - 2ln(x) = 1
ln(x^2/x^2) = 1
x^2/x^2 = e^1
1 = e
Therefore, the particle needs to pass through the point (1, e^1) or (1, e) for the rate of change of the distance to be 2 units per second.
To find the rate at which the particle is moving away from the origin, we need to calculate the rate of change of the distance between the particle and the origin as it moves along the curve y = ln(x).
Let's call the distance between the particle and the origin as d. We want to find d' (the derivative of d with respect to time t) when the particle passes through the point (1, e).
We can use the distance formula to calculate d:
d = sqrt((x - 0)^2 + (ln(x) - 0)^2)
= sqrt(x^2 + ln(x)^2)
Taking the derivative of d with respect to x, we get:
d' = (d/dx) sqrt(x^2 + ln(x)^2)
Using the chain rule, we can simplify the derivative expression:
d' = (1/2) * (x^2 + ln(x)^2)^(-1/2) * (2x + 2ln(x) * (1/x))
Now we know that the abscissa x is increasing at a rate of 2 units per second. So, dx/dt = 2.
Substituting x = 1 and dx/dt = 2 into the expression for d', we can calculate the rate at which the particle is moving away from the origin:
d' = (1/2) * (1^2 + ln(1)^2)^(-1/2) * (2 * 1 + 2ln(1) * (1/1))
d' = (1/2) * (1 + 0)^(-1/2) * (2 + 0)
d' = (1/2) * 2
d' = 1
Therefore, the particle is moving away from the origin at a rate of 1 unit per second as it passes through the point (1, e).