Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,459 questions

17,854 answers

1,446 comments

808,042 users

Find the points on the curve where the tangent is horizontal or vertical.

0 votes
Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to check your work.

x = cos θ,         y = cos 3θ
asked Feb 2, 2015 in CALCULUS by anonymous

2 Answers

0 votes

Step 1:

The curve is , .

Find the points on the curve where the tangent line is horizontal or vertical.

Slope of the horizontal tangent line is 0.

Slope of the vertical tangent line is .

Find the slope of the curve.

Consider .

Differentiate on each side with respect to .

Consider .

Differentiate on each side with respect to .

Step 2:

Slope of the tangent line is first derivative of the curve.

Slope of the curve is

Substitute  and in the above equation.

Slope of the tangent line is .

answered Feb 7, 2015 by lilly Expert
0 votes

Contd......

Step 3:

Slope of the horizontal tangent line is 0.

Consider .

image

Consider image.

image

Now substitute image in and  .

image

Now substitute image in and .

image

The points on the curve where the tangent line is horizontal are image and image.

Step 4:

Slope of the vertical tangent line is.

image

Since the slope is not defined, there is no vertical tangent lines to the given curve.

Step 5:

To check the result, graph the curve.

The graph of the curve , is :

image

Observe the graph of the curve notice that, the curve has horizontal tangent lines at the points image and image.

Solution:

The points on the curve where the tangent line is horizontal are image and image.

There is no vertical tangent lines to the given curve.

answered Feb 7, 2015 by lilly Expert

Related questions

asked Feb 2, 2015 in CALCULUS by anonymous
asked Jan 20, 2015 in CALCULUS by anonymous
asked Jan 20, 2015 in CALCULUS by anonymous
...