Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.
x = 5 ln t, y = 8sqrt(t), z = t^5
(0,8,1)
(x(t),y(t),z(t))=( )
see your text for examples, or
http://mathinsight.org/parametrized_curve_tangent_line_examples
To find the parametric equations for the tangent line to the curve at the specified point, we need to find the derivatives of the given equations with respect to the parameter t. Then, we substitute the values of t and plug them into the equations of the tangent line.
Start by finding the derivatives of the given equations:
dx/dt = 5/t
dy/dt = (4/t^(1/2))
dz/dt = 5t^4
Next, we need to find the values of t that correspond to the point (0,8,1). To determine the value of t, we set each coordinate equation equal to the corresponding coordinate of the point and solve for t.
For x=0, we have:
5 ln t = 0
ln t = 0
t = 1
For y=8, we have:
8√t = 8
√t = 1
t = 1
For z=1, we have:
t^5 = 1
t = 1^(1/5)
t = 1
Thus, the value of t at the point (0,8,1) is t = 1.
Now that we have the value of t, we can find the derivatives at that point:
dx/dt = 5/1 = 5
dy/dt = 4/√1 = 4
dz/dt = 1^5 = 1
So, the derivatives at t=1 are dx/dt = 5, dy/dt = 4, dz/dt = 1.
Finally, we can write the parametric equation of the tangent line using the point (0,8,1) and the derivatives dx/dt, dy/dt, and dz/dt at t=1:
x(t) = x(1) + (dx/dt)(t - 1)
y(t) = y(1) + (dy/dt)(t - 1)
z(t) = z(1) + (dz/dt)(t - 1)
Plugging in the values, we get:
x(t) = 0 + 5(t - 1)
y(t) = 8 + 4(t - 1)
z(t) = 1 + 1(t - 1)
Simplifying these equations, we get:
x(t) = 5t - 5
y(t) = 4t + 4
z(t) = t
Therefore, the parametric equations for the tangent line to the curve at the point (0,8,1) are:
x(t) = 5t - 5
y(t) = 4t + 4
z(t) = t