Find the parametric equations for the tangent line to the curve with the given parametric equations at specified point.
x= e^t
y=te^t
z=te^(t^2)
(1,0,0)
To find the parametric equations for the tangent line, we will first find the vectors representing the tangent lines at the given point in each direction, which are obtained by the derivative of the position vector with respect to the parameter.
dx/dt = d(e^t)/dt = e^t
dy/dt = d(te^t)/dt = e^t + te^t
dz/dt = d(te^(t^2))/dt = e^(t^2) + 2t^2 * e^(t^2)
Now, we plug in the given point (1,0,0), which corresponds to t = 0:
dx/dt|_(t=0) = 1
dy/dt|_(t=0) = 1
dz/dt|_(t=0) = 1
So the tangent vector is (1,1,1).
The parametric equations for the tangent line are:
x = 1 + t
y = 0 + t
z = 0 + t
To find the parametric equations for the tangent line to the curve at a specified point, we need to find the derivative of each equation x(t), y(t), and z(t) with respect to parameter t.
Let's start by finding the derivative of x(t) = e^t:
dx/dt = d/dt(e^t) = e^t
Next, let's find the derivative of y(t) = te^t:
dy/dt = d/dt(te^t) = e^t + t(e^t)
Now, find the derivative of z(t) = te^(t^2):
dz/dt = d/dt(te^(t^2)) = e^(t^2) + 2t^2e^(t^2)
Now we have the derivatives of each component of the curve.
Next, we need to find the value of t that corresponds to the given point (1,0,0). We want to find the t-value that makes x(t) = 1, y(t) = 0, and z(t) = 0.
From x(t) = e^t = 1, we can take the natural logarithm of both sides to get t = ln(1) = 0.
From y(t) = te^t = 0, we can see that t = 0 satisfies this equation.
Similarly, from z(t) = te^(t^2) = 0, we can see that t = 0 satisfies this equation as well.
Therefore, the point on the curve that corresponds to t = 0 is (1, 0, 0).
Now, we have the point (1, 0, 0) and the derivatives dx/dt, dy/dt, and dz/dt at t = 0.
The parametric equations for the tangent line to the curve at this point can be obtained by using the point-normal form of the equation of a line.
x - x0 / dx/dt = y - y0 / dy/dt = z - z0 / dz/dt
Plugging in the values, we have:
(x - 1) / e^t = y / (e^t + t(e^t)) = z / (e^(t^2) + 2t^2e^(t^2))
Since t = 0, we can simplify this equation:
(x - 1) / 1 = y / 1 = z / 1
Therefore, the parametric equations for the tangent line to the curve at the point (1, 0, 0) are:
x = t + 1
y = t
z = t