a) A curve is defined by the following parametric equations
x= 3t/t^2 +1 , y=1/t^2 +1
(i) Find an expression, in terms of t for the gradient of the curve.
(ii) Determine the value of the gradient at the point where t=0
To find the gradient of the curve, we need to differentiate both x and y with respect to t.
(i) Differentiating x with respect to t:
dx/dt = (3(1)(t^2 + 1) - 3t(2t))/(t^2 + 1)^2
= (3t^2 + 3 - 6t^2)/(t^2 + 1)^2
= (-3t^2 + 3)/(t^2 + 1)^2
Differentiating y with respect to t:
dy/dt = (2t)(1) / (t^2 + 1)^2
= 2t / (t^2 + 1)^2
Therefore, the expression for the gradient of the curve is:
dy/dx = (dy/dt)/(dx/dt)
= (2t / (t^2 + 1)^2) / (-3t^2 + 3)/(t^2 + 1)^2
= 2t / (-3t^2 + 3)
= 2t / 3(1 - t^2)
(ii) To determine the value of the gradient at the point where t = 0, substitute t = 0 into the expression for the gradient:
dy/dx = 2t / 3(1 - t^2)
dy/dx = 2(0) / 3(1 - (0)^2)
dy/dx = 0 / 3(1 - 0)
dy/dx = 0
Therefore, the value of the gradient at the point where t = 0 is 0.
(i) To find the expression for the gradient of the curve, we need to find the derivative dy/dx.
First, let's express y in terms of x by substituting the given parametric equations:
x = 3t / (t² + 1) (1)
y = 1 / (t² + 1) (2)
Now, we can differentiate equation (1) with respect to t:
dx/dt = (3(t² + 1) - 3t(2t)) / (t² + 1)²
=> dx/dt = (3t² + 3 - 6t²) / (t² + 1)²
=> dx/dt = (-3t² + 3) / (t² + 1)²
Next, we can find dy/dt by differentiating equation (2) with respect to t:
dy/dt = (-2t) / (t² + 1)²
Finally, to find dy/dx, we can use the chain rule:
dy/dx = (dy/dt) / (dx/dt)
=> dy/dx = ((-2t)/(t² + 1)²) / ((-3t² + 3) / (t² + 1)²)
=> dy/dx = (-2t) / (-3t² + 3)
=> dy/dx = 2t / (3 - 3t²)
So, the expression for the gradient of the curve is given by dy/dx = 2t / (3 - 3t²).
(ii) To determine the value of the gradient at the point where t = 0, we substitute t = 0 into the expression for the gradient:
dy/dx = 2(0) / (3 - 3(0)²)
=> dy/dx = 0 / 3
=> dy/dx = 0
Therefore, the value of the gradient at the point where t = 0 is 0.
Assuming the usual carelessness with parentheses, I assume you mean
x = 3t/(t^2+1)
y = 1/(t^2+1)
Now, the gradient is just dy/dx = (dy/dt) / (dx/dt)
dy/dt = -2t/(t^2+1)^2
dx/dt = -3(t^2-1)/(t^2+1)^2
so,
dy/dx = -2t/(t^2+1)^2 * (t^2+1)^2/(-3(t^2-1) = 2t/(3(t^2-1))
at t=0, dy/dx = 0