Finding a Tangent Line

• Oct 21st 2012, 12:13 PM
smgrojean
Finding a Tangent Line
How can I go about finding a tangent line for the following:

(x^3)*(R^5)=1 at the point (x,R)=(1,1)

I took the derivative of the equation but I didn't know where to go after that.
• Oct 21st 2012, 12:40 PM
skeeter
Re: Finding a Tangent Line
$\frac{dR}{dx}$ is the slope of the curve at any point $(x,R)$ on the curve ...

evaluate the derivative at $x = 1$ , $R = 1$ to find the slope of the tangent line at that specific point, then use the point-slope form of a linear equation ...

$R - R_1 = m(x - x_1)$
• Oct 23rd 2012, 03:28 PM
smgrojean
Re: Finding a Tangent Line
So if I take the derivative I get:

3*R^5*X^2

Then if I evaluate that at x=1, R=1 I get:

3*1^5*1^2 = 3

So the point slope of the linear equation should be the following:

R - 1 = 3 ( X - 1)
• Oct 23rd 2012, 05:10 PM
Prove It
Re: Finding a Tangent Line
That is not the derivative. Since R is a function of x, you will need to use implicit differentiation.
• Oct 23rd 2012, 05:25 PM
skeeter
Re: Finding a Tangent Line
$x^3 \cdot R^5 = 1$

solve for $R$ in terms of $x$ ...

$R = x^{-\frac{3}{5}}$

... now find $\frac{dR}{dx}$
• Oct 26th 2012, 12:37 PM
smgrojean
Re: Finding a Tangent Line
so the derivative would actually be:

dr/dx = -3/(5*x^(8/5))
• Oct 26th 2012, 01:53 PM
skeeter
Re: Finding a Tangent Line
ok ... find the equation of the tangent line.
• Oct 28th 2012, 11:26 AM
smgrojean
Re: Finding a Tangent Line
so the equation should be:

R - 1 = -3/5(x-1)