An object is moving along the graph of f (x)= x2 . When it reaches the point (2,4) the x coordinate of the object is decreasing at the rate of 3 units/sec. Give the rate of change of the distance between the object and the point (0,1) at the instant when the object is at (2,4).
Let P(x,y) be any point on f(x)
then the distance D between P and (0,1) is such that
D^2 = (x-0)^ + (y-1)^2
= x^2 + (x^2 - 1)^2
= x^2 + x^4 - 2x^2 + 1 = x^4 - x^2 + 1
2D dD/dt = 4x^3 dx/dt - 2x dx/dt
so at (2,4), x=2 and dx/dt = -3
and D^2 = 4 + 9 = 13
D = √13
√13 dD/dt = 4(8)(-3) - 2(2)(-3)
dD/dt = -84/√13 = appr. -23.3 units/sec
check my arithmetic.
OK until you forgot that it was 2D dD/dt, and you just plugged in √13. So, the answer is half what you showed.
To find the rate of change of the distance between the object and the point (0,1) at the instant when the object is at (2,4), we need to use the distance formula.
The distance between two points (x₁, y₁) and (x₂, y₂) is given by the formula:
d = √((x₂ - x₁)² + (y₂ - y₁)²)
In this case, the object is at (2,4) and the point is at (0,1). Let's label the object's x-coordinate as x(t) and the corresponding y-coordinate as y(t), where t represents time. We're given that the object is moving along the graph of f(x) = x², so the object's position can be described as (x(t), y(t)) = (x(t), f(x(t)))) = (x(t), x(t)²).
At the instant when the object is at (2,4), the x-coordinate of the object is decreasing at a rate of 3 units/sec. This means dx/dt = -3.
To find the rate of change of the distance, we need to differentiate the distance formula with respect to time:
d/dt(d) = d/dt(√((x(t) - 0)² + (f(x(t)) - 1)²))
Using the chain rule, we have:
d/dt(d) = ((x(t) - 0) * dx/dt) / √((x(t) - 0)² + (f(x(t)) - 1)²) + ((f(x(t)) - 1) * (df/dx * dx/dt))
Since f(x) = x², df/dx = 2x.
Substituting the given values:
d/dt(d) = (2 * dx/dt) / √((x(t) - 0)² + (f(x(t)) - 1)²) + ((f(x(t)) - 1) * (2x * dx/dt))
At the point (2,4), we have x(t) = 2 and f(x(t)) = f(2) = 4.
Substituting these values and dx/dt = -3 into the equation:
d/dt(d) = (2 * (-3)) / √((2 - 0)² + (4 - 1)²) + ((4 - 1) * (2 * (-3)))
Simplifying further:
d/dt(d) = (-6) / √((2 - 0)² + (4 - 1)²) + (3 * (-6))
Calculating √(2² + 3²):
d/dt(d) = (-6) / √(4 + 9) + (-18)
d/dt(d) = (-6) / √13 - 18
Therefore, the rate of change of the distance between the object and the point (0,1) at the instant when the object is at (2,4) is (-6) / √13 - 18 units/sec.