Parametric equation of tangent line

[jwxie]

Homework Statement

Find parametric equations for the tangent line to the curve with the given parametric equations at a given point.
\[x = t^5, y = t^4, z = t^3\] at point (1,1,1)

Homework Equations

The Attempt at a Solution

So we need to have direction vector, and a point.
To find the tangent vector, we may get it through taking \[\frac{\mathrm{dt} }{\mathrm{d} x,y,z}\] respectively.

So I get \[r^{'}(t) = <5t^4, 4t^3, 3t^2>\].

In the end, using the formula
\[r(t) = r_{point} + t*(r^{'}(t))\]

Putting together, I get
x = 5t^5 + 1
y = 4t^5 + 1
y = 3t^5 + 1

But the book gives
x = 5t + 1
y = 4t + 1
y = 3t + 1

What is my mistake?
Thank you for any input!

 

[ZioX]

The book evaluated r'(t) at the point (1,1,1) whereas you didn't.

 

[jwxie]

Is it still wrong though?
I was reading the note, and apparently my professor forgot to evaluate r'(t) at the point too.

 

[HallsofIvy]

Yes, if you don't do what is asked, then it is wrong!

By the way, you did NOT "take \[\frac{\mathrm{dt} }{\mathrm{d} x,y,z}\]
respectively", you took \frac{dx}{dt}, \frac{dy}{dt}, and \frac{dz}{dt}[/itex]- although it might have been better to simply say that you differentiated the &quot;position vector&quot;, t^5\vec{i}+ t^4\vec{j}+ t^3\vec{k} with respect to t to get the tangent vector 5t^4\vec{i}+ 4t^3\vec{j}+ 3t^2\vec{k}. At the given point, (1, 1, 1), where t= 1, that is 5\vec{i}+ 4\vec{j}+ 3\vec{k} and those become the coefficients for the parameter: x= 5t+ 1, y= 4t+ 1, z= 3t+ 1.<br /> <br /> Since we have no idea what &quot;note&quot; you are reading we cannot say whether you professor forgot to evaluate at the given point. Was the result a <b>line</b>? If so, then he could not have.<br /> <br /> Obviously, the equations for a tangent <b>line</b> must be <b>linear</b> which your <br /> x = 5t^5 + 1<br /> y = 4t^5 + 1<br /> y = 3t^5 + 1<br /> is not.