Answer:
Explanation:
a. To find the rate at which the x-coordinate of the bug is increasing when the bug is at the point (4, 16), we need to differentiate the equation y=x with respect to time t:
dy/dt = dx/dt
Since the bug is moving along the right side of the parabola y=x, the bug's position can be described by the equation y=x^2. Taking the derivative of both sides with respect to time t, we get:
2y(dy/dt) = 2x(dx/dt)
Simplifying and plugging in the given values for y and dy/dt:
2(16)(9) = 2(4)(dx/dt)
dx/dt = 36 cm/min
Therefore, the x-coordinate of the bug is increasing at a rate of 36 cm/min when the bug is at the point (4, 16).
b. The equation y=x^2 can be rewritten as x=sqrt(y). Differentiating both sides with respect to time t, we get:
dx/dt = 1/(2sqrt(y)) * dy/dt
Substituting y=16 and dy/dt=9, we get:
dx/dt = 1/(2sqrt(16)) * 9
dx/dt = 9/8 cm/min
c. We can use the same equation from part (a) to find the rate at which the y-coordinate of the bug is increasing when the bug is at the point (4, 16):
2y(dy/dt) = 2x(dx/dt)
Substituting y=16, x=4, and dx/dt=36, we get:
2(16)(dy/dt) = 2(4)(36)
Solving for dy/dt:
dy/dt = 18 cm/min
Therefore, the y-coordinate of the bug is increasing at a rate of 18 cm/min when the bug is at the point (4, 16).
d. Let D be the distance the bug is from the origin. We can use the Pythagorean theorem to relate D to x and y:
D^2 = x^2 + y^2
Substituting y=x^2, we get:
D^2 = x^2 + x^4
Taking the derivative of both sides with respect to time t, we get:
2D(dD/dt) = 2x(dx/dt) + 4x^3(dx/dt)
Simplifying and substituting x=4, dx/dt=36, and y=16:
2sqrt(16+16^2)(dD/dt) = 2(4)(36) + 4(4^3)(36)
Solving for dD/dt:
dD/dt = (836 + 44^3*36) / (2sqrt(16+16^2))
dD/dt = 72/(sqrt(17))
Therefore, the distance between the bug and the origin is increasing at a rate of 72/(sqrt(17)) cm/min when the bug is at the point (4, 16).