# Paramentric eq. of tangent line

• Calcgeek123
In summary: Then once you find the slope, you can find the equation of the line.In summary, the conversation discusses finding the equation of the tangent line to a given curve at a specific point. The process involves finding the slope of the curve using the derivative and then using the slope to find the equation of the line. The conversation also touches on simplifying equations and using a point on the curve to calculate the slope.
Calcgeek123

## Homework Statement

Find the equation of the tangent line to f(t)=<cot(t),csc(t)> at the point (1/sq.rt of 3, 2/sq.rt of 3)

n/a

## The Attempt at a Solution

I started by finding the slope, y'/x', so I got csc(t)cot(t)/csc^2(t). I then used the equation of a line, y=mx+b. Doesn't a tangent line have the opposite slope? So I think i'd plug in 2/sq.rt of 3 for y, csc^2(t)/csc(t)cot(t) for the slope, 1/sq.rt of 3 for x, and then solve for b. Do I need to plug in my x and y points though into the slope? Then that gets tough because I don't know what csc^2(1/sq. rt of 3) is. Any suggestions?

Calcgeek123 said:

## Homework Statement

Find the equation of the tangent line to f(t)=<cot(t),csc(t)> at the point (1/sq.rt of 3, 2/sq.rt of 3)

n/a

## The Attempt at a Solution

I started by finding the slope, y'/x', so I got csc(t)cot(t)/csc^2(t).

Any law against simplifying that?

I then used the equation of a line, y=mx+b.

Why use that form instead of the form y - y0 = m(x - x0)? After all, you are given a point on the curve through which the line must pass.
Doesn't a tangent line have the opposite slope?

Opposite? Isn't the slope of the tangent line defined to be the slope of the curve?

So I think i'd plug in 2/sq.rt of 3 for y, csc^2(t)/csc(t)cot(t) for the slope, 1/sq.rt of 3 for x, and then solve for b. Do I need to plug in my x and y points though into the slope? Then that gets tough because I don't know what csc^2(1/sq. rt of 3) is. Any suggestions?

Can't you figure out what t gives the point in question and just use it in your calculations?

I say you should try to figure out what value of t makes that point. Once you do that you can derive that vector equation, then with the derivative you can find the slope at any given point. With that slope you can find the equation of the line.

Thank you! I took the advice from the first post, and I've ended up with y-csc(t)=cos(t) (x-csc(t)). Now I'm just lost as to what to do with the point I've been given, (1/sq. rt of 3, 2/sq. rt of 3). I could plug in those numbers into the equation i arrived at for x and y, but then what would i solve for?

do what i said on my post after that, because if you have the derived vector equation, you need to plug in a t value

## 1. What is the point-slope form of the parametric equation of a tangent line?

The point-slope form of the parametric equation of a tangent line is given by y-y0 = m(x-x0), where (x0, y0) is a point on the tangent line and m is the slope of the line.

## 2. How do you find the slope of a tangent line using parametric equations?

To find the slope of a tangent line using parametric equations, you can take the derivative of the parametric equations with respect to the parameter, t. The resulting expression will give you the slope of the tangent line at a specific point.

## 3. Can you find the equation of a tangent line using only parametric equations?

Yes, you can find the equation of a tangent line using only parametric equations. Once you have the slope and a point on the tangent line, you can use the point-slope form of the equation to find the equation of the line.

## 4. How do you determine if a point lies on a tangent line using parametric equations?

To determine if a point lies on a tangent line using parametric equations, you can plug the x and y values of the point into the parametric equations. If the resulting values satisfy the equations, then the point lies on the tangent line.

## 5. Can parametric equations be used to find the equation of a tangent line to a curve at a point that is not on the curve?

No, parametric equations can only be used to find the equation of a tangent line at a point that lies on the curve. If the point is not on the curve, then the tangent line cannot be determined using parametric equations.

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