A particle moves along the curve y=square sqrt 1+x^3. As it reaches the point (2,3), the y-coordinate is increasing ata a rate of 4cm/s. How fast is the x-coordinate of the point changing at that instant?
Well, well, well, it seems like we have a particle going for a joyride along a curve! Let's find out how fast it's moving, shall we?
To do that, we're gonna need to use the chain rule from calculus. Hold on tight, we're about to get mathy!
First, let's differentiate the given equation with respect to y to get it in terms of x: y = (1 + x^3)^1/2.
dy/dx = (1/2)(1 + x^3)^-1/2 * 3x^2
Now, we know that when the particle reaches the point (2,3), the y-coordinate is increasing at a rate of 4 cm/s. So, we can plug in those values and solve for dx/dt:
4 = (1/2)(1 + 2^3)^-1/2 * 3(2)^2 * dx/dt
Simplifying that mess, we get:
4 = (1/2)(9)^-1/2 * 12 * dx/dt
Now, divide by all the constants and solve for dx/dt:
dx/dt = 4 / [(1/2)(9)^-1/2 * 12]
Okay, my dear friend, it's time to crunch some numbers. Grab your trusty calculator and let's compute that monstrosity!
dx/dt ≈ 4 / 1.069
So, approximately, the x-coordinate is changing at a rate of 3.7 cm/s. Time to fasten your seatbelt, because this particle is on the move!
Hope that helps, and remember, always buckle up, even in the wacky world of mathematics!
To find the rate at which the x-coordinate (dx/dt) is changing, we can use implicit differentiation. Starting with the equation y = √(1+x^3), we differentiate both sides with respect to time (t):
d/dt(y) = d/dt(√(1+x^3))
To differentiate the left side, we use the chain rule:
dy/dt = d/dt(√(1+x^3))
dy/dt = (1/2)(1+x^3)^(-1/2) * d/dt(1+x^3)
Since we are given that dy/dt = 4 cm/s, we can substitute this value in:
4 = (1/2)(1+2^3)^(-1/2) * d/dt(1+2^3)
Simplifying:
4 = (1/2)(1+8)^(-1/2) * d/dt(9)
4 = (1/2)(9)^(-1/2) * d/dt(9)
Evaluating the square roots:
4 = (1/2)(1/3) * d/dt(9)
4 = 1/6 * d/dt(9)
Multiplying both sides by 6:
24 = d/dt(9)
Therefore, the x-coordinate of the point is changing at a rate of 24 cm/s at that instant.
To find how fast the x-coordinate is changing at the given instant, we can use the chain rule from calculus. The chain rule states that if y is a function of u and u is a function of x, then dy/dx = dy/du * du/dx.
In this case, we are given the rate of change of the y-coordinate, so we need to calculate dy/dx. First, let's find the derivative of y with respect to x (dy/dx).
Given: y = sqrt(1 + x^3)
To differentiate this equation, we can use the chain rule as follows:
dy/dx = (dy/du) * (du/dx)
Let's find each part step-by-step:
1. Find the derivative of y (dy/du):
For y = sqrt(1 + x^3), we can rewrite it as y = (1 + x^3)^(1/2). Now, differentiate using the power rule:
dy/du = (1/2)*(1 + x^3)^(-1/2) * (d(1 + x^3)/dx)
2. Find the derivative of (1 + x^3) with respect to x (d(1 + x^3)/dx):
Using the power rule, the derivative of x^3 is 3x^2. So, d(1 + x^3)/dx = d(1)/dx + d(x^3)/dx = 0 + 3x^2 = 3x^2.
Now, combinine the two parts to find dy/dx:
dy/dx = (dy/du) * (du/dx)
= (1/2)*(1 + x^3)^(-1/2) * 3x^2
Now we can find how fast the x-coordinate is changing.
Given: dy/dx = (1/2)*(1 + x^3)^(-1/2) * 3x^2
At the point (2,3), we know that dy/dx = 4 cm/s.
Substituting the values into the equation, we get:
4 = (1/2)*(1 + 2^3)^(-1/2) * 3*(2^2)
Simplifying the equation further gives:
4 = (1/2)*(1 + 8)^(-1/2) * 3*4
Now, solve for the x-coordinate:
x-coordinate at (2,3) = 2
Hence, we have:
4 = (1/2)*(9)^(-1/2) * 12
Now, cross-multiply and solve for the rate of change of the x-coordinate:
4*(2^2) = (1/2)*(9)^(-1/2) * dx/dt
16 = (1/2)*(9)^(-1/2) * dx/dt
Solve for dx/dt:
dx/dt = 16 / [(1/2)*(9)^(-1/2)]
Hence, the rate at which the x-coordinate is changing at that instant is dx/dt = 16 / [(1/2)*(9)^(-1/2)] cm/s.