# Thread: Need Help (tangent to parameterized curves)

1. ## Need Help (tangent to parameterized curves)

I have 2 questions:

The first says: find an equation for the line tangent to the ccurve at the point defined by the given value of t. Also find the value of d^2y/dx^2.

x= t y= sq root(t) t= 1/4

(I am not sure how to do this problem so an explanation would be great)

Also I have a question that says: The position of a particle moving along a coordinate line is s= sq root (1+4t) with s in meters and t inseconds. Find the particles velocity and acceleration at t=6

For this problem I just found the derivative which is

(1/2(1+4t)^-1/2) x (4t) and I got an answer of 6 for the velocity and the back of my book has 2/5 as the answer. What am I doing wrong?

2. Originally Posted by KarlosK
I have 2 questions:

The first says: find an equation for the line tangent to the ccurve at the point defined by the given value of t. Also find the value of d^2y/dx^2.

x= t y= sq root(t) t= 1/4

(I am not sure how to do this problem so an explanation would be great)

Also I have a question that says: The position of a particle moving along a coordinate line is s= sq root (1+4t) with s in meters and t inseconds. Find the particles velocity and acceleration at t=6

For this problem I just found the derivative which is

(1/2(1+4t)^-1/2) x (4t) and I got an answer of 6 for the velocity and the back of my book has 2/5 as the answer. What am I doing wrong?
(1) for a tangent line, you need two things ... the point of tangency $(x,y)$and the slope at that point , $\frac{dy}{dx}
$
.

the point comes straight from evaluating $x(t)$ and $y(t)$ at the desired value of $t$.

slope ... $m = \frac{dy}{dx} = \frac{y'(t)}{x'(t)}$

for the 2nd derivative, note that $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$

$\frac{d}{dx}\left(\frac{y'(t)}{x'(t)}\right) = \frac{x'(t) \cdot y''(t) \cdot \frac{dt}{dx} - y'(t) \cdot x''(t) \cdot \frac{dt}{dx}}{[x'(t)]^2} = \frac{x'(t) \cdot y''(t) - y'(t) \cdot x''(t)}{[x'(t)]^3}$

$s = \sqrt{1+4t} = (1+4t)^{\frac{1}{2}}
$

$v = s'(t) = \frac{1}{2}(1+4t)^{-\frac{1}{2}} \cdot 4 = \frac{2}{\sqrt{1+4t}}$

$v(6) = \frac{2}{\sqrt{25}} = \frac{2}{5}$

note that $a = v'(t) = s''(t)$