A point moves along the curve y=x^2+1 so that the x-coordinate increases at a constant rate of 5 units per second. When x=1, at what rate is the gradient of the curve increasing?
Thanks in advance to anyone who helps. Answer is 3
Sorry is meant to say:
A point moves along the curve y=x^3 so that the x-coordinate increases at a constant rate of 5 units per second. When x=1, at what rate is the gradient of the curve increasing?
dx/dt = 5
dy/dt = 3x^2 dx/dt
Now, the gradient is g(x) = dy/dx = 3x^2
the rate of change of the gradient is dg/dt = 6x dx/dt = 6*5 = 30
If the answer is supposed to be 3, then I must have misread the problem.
O course, there's always the chance that you mis-stated it ...
To find the rate at which the gradient of the curve is increasing, we can differentiate the equation of the curve with respect to x.
Taking the derivative of y = x^2 + 1 with respect to x, we get:
dy/dx = 2x
This represents the gradient of the curve at any given point (x, y) on the curve.
Now, we are given that the x-coordinate is increasing at a constant rate of 5 units per second, so dx/dt = 5.
To find the rate at which the gradient is increasing, we need to differentiate dy/dx with respect to time t. Using the chain rule, we have:
d(dy/dx)/dt = d(dy/dx)/dx * dx/dt
Since dx/dt is a constant (5), its derivative is 0. Therefore, the expression simplifies to:
d(dy/dx)/dt = (d^2y/dx^2) * (dx/dt)
Now, let's find the value of (d^2y/dx^2) at x = 1:
To do this, we differentiate dy/dx = 2x with respect to x:
(d^2y/dx^2) = d(2x)/dx = 2
Substituting the values in the equation for d(dy/dx)/dt:
d(dy/dx)/dt = (d^2y/dx^2) * (dx/dt) = 2 * 5 = 10
Therefore, at x = 1, the rate at which the gradient of the curve is increasing is 10.
To find the rate at which the gradient of the curve is increasing, we need to first find the derivative of the curve equation with respect to x, and then determine the rate of change of the derivative with respect to time.
Given curve equation: y = x^2 + 1
Step 1: Find the derivative of the curve equation with respect to x.
To find the derivative, we use the power rule for differentiation. Since the equation is y = x^2 + 1, we differentiate term by term:
dy/dx = d/dx (x^2) + d/dx (1)
Differentiating x^2 with respect to x gives us:
dy/dx = 2x + 0
dy/dx = 2x
Step 2: Determine the rate of change of the derivative (dy/dx) with respect to time.
We know that the x-coordinate is increasing at a constant rate of 5 units per second. This means dx/dt = 5.
Applying the chain rule, we can express the rate of change of y (dy/dt) in terms of dx/dt:
dy/dt = (dy/dx) * (dx/dt)
Substituting the values we know:
dy/dt = (2x) * (dx/dt)
dy/dt = (2x) * 5
dy/dt = 10x
Step 3: Determine the rate of change of the gradient when x = 1.
To find the rate of change of the gradient when x = 1, substitute x = 1 into the expression derived in the previous step:
dy/dt = 10(1)
dy/dt = 10
Therefore, when x = 1, the rate at which the gradient of the curve is increasing is 10.