Can someone tell me how to do this type of problem? A particle moves along the curve y=sqrt(1+x^3). As it reaches the point (2,3) the y coorindate is increasing at a rate os 4cm/s. How fast is the x coordinate of the point changing at that instant?
Differentiate both sides with respect to t:
dy/dt = d(sqrt(1+x^3))/dx * dx/dt
=(3x²/(2sqrt(1+x³)))*dx/dt
So given x=2,y=3 and dy/dt=3 cm/s
substitute in formula above to solve for dx/dt.
I get dx/dt=2.
To solve this problem, we need to use related rates. Related rates problems involve finding the rate at which one quantity is changing while knowing the rates at which other related quantities are changing.
Let's first identify the given information:
- The curve is defined by the equation y = sqrt(1 + x^3)
- The particle is at the point (2, 3)
- The y-coordinate is increasing at a rate of 4 cm/s
We need to find how fast the x-coordinate is changing at the instant when the particle is at (2, 3).
To solve this problem, we can use differentiation.
1. Take the derivative of both sides of the equation to relate dx/dt and dy/dt:
dy/dt = d(sqrt(1 + x^3))/dt
2. Differentiate the right side using chain rule and the power rule:
dy/dt = [(1 + x^3)^(1/2)]' * dx/dt
= (1/2)(1 + x^3)^(-1/2) * 3x^2 * dx/dt
= 3x^2 / [2sqrt(1 + x^3)] * dx/dt
3. Substitute the given values into the equation:
- Let x = 2 (from the point (2, 3))
- Let dy/dt = 4 cm/s
4 = 3(2^2) / [2sqrt(1 + 2^3)] * dx/dt
4. Solve for dx/dt:
8 = 12 / [2sqrt(9)] * dx/dt
8 = 6 / [sqrt(9)] * dx/dt
8 = 6/3 * dx/dt
8 = 2 * dx/dt
dx/dt = 8/2
dx/dt = 4 cm/s
Thus, the x-coordinate of the point is changing at a rate of 4 cm/s at the instant when the particle is at (2, 3).