160k views
1 vote
A bug is moving along the right side of the parabola y=x^2 at a rate such that its distance from the origin is increasing at 4 cm / min.

a. At what rate is the x-coordinate of the bug increasing when the bug is at the point (6. 36)?
b. Use the equation y=x^2 to find an equation relating to dy/dt to dx/dt
c. At what rate is the y-coordinate of the bug increasing when the bug is at the point (6, 36)?

1 Answer

5 votes

To solve this problem, we'll use the concept of related rates. Let's break down each part of the problem:

a. At what rate is the x-coordinate of the bug increasing when the bug is at the point (6, 36)?

Let's assume that the bug's x-coordinate is x, and its y-coordinate is y. Since the bug is moving along the right side of the parabola y = x^2, we have the equation y = x^2. We are given that the distance between the bug and the origin (which is √(x^2 + y^2)) is increasing at a rate of 4 cm/min. We need to find the rate at which the x-coordinate of the bug is changing, which is dx/dt.

Using the Pythagorean theorem, we have:

√(x^2 + y^2) = √(x^2 + (x^2)^2) = √(x^2 + x^4)

Differentiating both sides of the equation with respect to time (t), we get:

(d/dt)√(x^2 + x^4) = (d/dt)4

Applying the chain rule, we have:

(1/2) * (x^2 + x^4)^(-1/2) * (2x + 4x^3 * dx/dt) = 0

Simplifying, we get:

x + 2x^3 * dx/dt = 0

Substituting the coordinates of the bug at the given point (6, 36), we have:

6 + 2(6)^3 * dx/dt = 0

Solving for dx/dt, we get:

2(6)^3 * dx/dt = -6

dx/dt = -6 / (2(6)^3)

dx/dt = -1 / 72 cm/min

Therefore, the x-coordinate of the bug is decreasing at a rate of 1/72 cm/min when the bug is at the point (6, 36).

b. Use the equation y = x^2 to find an equation relating dy/dt to dx/dt

We can differentiate the equation y = x^2 with respect to time (t) using the chain rule:

(d/dt)(y) = (d/dt)(x^2)

dy/dt = 2x * dx/dt

Using the equation y = x^2, we can substitute x = √y into the equation above:

dy/dt = 2√y * dx/dt

This equation relates the rate of change of y (dy/dt) to the rate of change of x (dx/dt) for points on the parabola y = x^2.

c. At what rate is the y-coordinate of the bug increasing when the bug is at the point (6, 36)?

To find the rate at which the y-coordinate of the bug is increasing, we need to determine dy/dt.

Using the equation derived in part b, we have:

dy/dt = 2√y * dx/dt

Substituting the given values at the point (6, 36), we have:

dy/dt = 2√36 * (-1/72)

Simplifying, we get:

dy/dt = -2/72

dy/dt = -1/36 cm/min

Therefore, the y-coordinate of the bug is decreasing at a rate of 1/36 cm/min when the bug is at the point (6, 36).

User Dheeraj Palagiri
by
8.5k points

No related questions found