A particle is moving along the curve below.
y = sqrt(x)
As the particle passes through the point (4,2), its x-coordinate increases at a rate of 4 cm/s. How fast is the distance from the particle to the origin changing at this instant?
Well, we know x=3 and y=8 and dx/dt=4. We need to find z (the distance from the particle to the origin) dy/dt, and finally our answer dz/dt.
First solve for dy/dt
y=2(3x+7)^(1/2)
dy/dt=[2(1/2)(3x+7)^(-1/2)](3dx/dt)
dy/dt=[(3x+7)^(-1/2)]3dx/dt
dy/dt=[(3(3)+7)^(-1/2)](3(4))
dy/dt=3
Now we can draw a right triangle, with x=3 and y=8, and using the pythagorean theorem, z=8.544
So we have z^2=x^2+y^2
(2z)dz/dt=(2x)dx/dt+(2y)dy/dt
(z)dz/dt=(x)dx/dt+(y)dy/dt
(8.544)dz/dt=(3)(4)+(8)(3)
dz/dt=4.2135
A particle is moving along the curve y = 2 √{3 x + 7}. As the particle passes through the point (3, 8), its x-coordinate increases at a rate of 4 units per second. Find the rate of change of the distance from the particle to the origin at this instant.
To find how fast the distance from the particle to the origin is changing at a specific point, we can use the concept of the derivative.
First, let's set up some variables:
Let x represent the x-coordinate of the particle.
Let y represent the y-coordinate of the particle.
Let d represent the distance from the particle to the origin.
We are given that the particle is moving along the curve y = sqrt(x), and it passes through the point (4, 2). This means that at this particular point, x = 4 and y = 2.
We want to find how fast d is changing at this point, which can be represented by the derivative of d with respect to time (dt).
To set up the relationship between x, y, and d, we can use the Pythagorean theorem:
d^2 = x^2 + y^2
Since we are looking for the rate of change of d with respect to time (dt), we can differentiate both sides of the equation with respect to t:
2d * dd/dt = 2x * dx/dt + 2y * dy/dt
Now, let's substitute the values we know:
At the point (4, 2), x = 4 and y = 2.
2d * dd/dt = 2(4) * 4 + 2(2) * dy/dt
Since we are interested in finding dy/dt, we need to isolate it:
2d * dd/dt - 8 * dx/dt = 4 * dy/dt
Now, we can solve for dy/dt:
dy/dt = (2d * dd/dt - 8 * dx/dt) / 4
Since the particle is moving along the curve y = sqrt(x), we know that at the point (4, 2), the x-coordinate is increasing at a rate of 4 cm/s (dx/dt = 4 cm/s). To find dd/dt, the rate at which the particle's distance from the origin is changing, we can substitute the values of x and y into the equation d^2 = x^2 + y^2:
d^2 = 4^2 + 2^2 = 16 + 4 = 20
Therefore, d = sqrt(20) = 2sqrt(5)
Substituting these values:
dy/dt = (2(2sqrt(5)) * dd/dt - 8 * 4) / 4
Simplifying:
dy/dt = (4sqrt(5) * dd/dt - 32) / 4
Finally, substituting the given value of dd/dt = 4 cm/s:
dy/dt = (4sqrt(5) * 4 - 32) / 4
dy/dt = (16sqrt(5) - 32) / 4
dy/dt = 4sqrt(5) - 8 cm/s
Therefore, the distance from the particle to the origin is changing at a rate of 4sqrt(5) - 8 cm/s at the instant the particle passes through the point (4, 2).
let P(x,y) be any point on the curve
y = x^(1/2)
dy/dt = (1/2)x^(-1/2) dx/dt
when x-4 and y=2 and dx/dt=4
dy/dt = (-1/2)(4^(-1/2)(4)
= ...
you finish it