Question

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = t, y = e−4t, z = 5t − t5; (0, 1, 0) x(t), y(t), z(t) = t,1−4t,5t Illustrate by graphing both the curve and the tangent line on a common screen.

Answer 1

Answer:

Step-by-step explanation:

At the point (0, 1,0) t = 0

Find the tangent vector:

$\frac{dx}{dt}= 1$

$\frac{dy}{dt}= -4e^{-4t}$

$\frac{dz}{dt}=5-5t^4$

The tangent vector for all points $\vec v(t)$ is

$\vec v(t) = \hat {i}-4e^{-4t}\hat{j}+(5-5t^4)\hat{k}$

$\rightarrow \vec v (0)= \hat{i}-4\hat{j}$

The vector equation of the tangent line is

$(x,y,z) = (0,1,0)+s(\hat{i}-4\hat{j})$

The parametric equation for this line are

$x= s$

$y=1-4s$

$z=0$

Answer 2

Parametric equations for the tangent line to the curve with the given parametric equations at the specified point are

x=s

y=1-4s

z=0

What is the parametric equation?

In mathematics, a parametric equation defines a group of quantities as functions of one or more independent variables called parameters.

At the point (0, 1,0) t = 0

Find the tangent vector:

$\dfrac{dx}{dt}=1$

$\dfrac{dy}{dt}=-4e^{-4t}$

$\dfrac{dz}{dt}=5-5t^4$

The tangent vector for all points  is

$v(t)=i-4e^{-4t}j+(5-5t^4)k$

$v(0)=i-4j$

The vector equation of the tangent line is

$(x,y,z)=(0,1,0)+s(i-4j)$

The parametric equation for this line is

$x=s\\\\y=1-4s\\\\z=0$

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