What is the equation of the tangent line of #r=-2sin(6theta-(4pi)/3) # at #theta=(2pi)/3#?

Answer 1

#y = -4(x-sqrt3/2)-1.5#

The reference Tangents with Polar Coordinates gives us the equation:

#dy/dx = ((dr)/(d theta)sin(theta)+rcos(theta))/((dr)/(d theta)cos(theta)-rsin(theta))" [1]"#

Compute #(dr)/(d theta)#

#(dr)/(d theta) = -12cos(6theta-(4pi)/3)#

#(dr)/(d theta) = 12sin(6theta+pi/6)#

Substitute right sides of the equation for r and #(dr)/(d theta)# into equation [1]:

#dy/dx = ((12sin(6theta+pi/6))sin(theta)+(-2sin(6theta-(4pi)/3))cos(theta))/((12sin(6theta+pi/6))cos(theta)-(-2sin(6theta-(4pi)/3))sin(theta))#

The slope, m, of the tangent line is the above equation evaluated at #theta=(2pi)/3#

#m = ((12sin(6(2pi)/3+pi/6))sin((2pi)/3)+(-2sin(6(2pi)/3-(4pi)/3))cos((2pi)/3))/((12sin(6(2pi)/3+pi/6))cos((2pi)/3)-(-2sin(6(2pi)/3-(4pi)/3))sin((2pi)/3))#

#m~~-4#

The x coordinate, #x_1#, of the point of tangency is:

#x_1 = -2sin(6(2pi)/3-(4pi)/3)cos((2pi)/3)#

#x_1 = sqrt3/2#

The y coordinate, #y_1#, of the point of tangency is:

#y_1 = -2sin(6(2pi)/3-(4pi)/3)sin((2pi)/3)#

#y_1 = -1.5#

Using the point-slope form of the equation of a line:

#y = m(x-x_1)+y_1#

We obtain the following equation:

#y = -4(x-sqrt3/2)-1.5#

Here is a graph of the function, the point of tangency, and the tangent line:

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the equation of the tangent line of ( r = -2\sin(6\theta - \frac{4\pi}{3}) ) at ( \theta = \frac{2\pi}{3} ), follow these steps:

  1. Calculate the derivative of ( r ) with respect to ( \theta ), denoted as ( \frac{dr}{d\theta} ).
  2. Evaluate ( \frac{dr}{d\theta} ) at ( \theta = \frac{2\pi}{3} ) to find the slope of the tangent line.
  3. Use the point-slope form of a line ( y - y_1 = m(x - x_1) ), where ( (x_1, y_1) ) is the point of tangency and ( m ) is the slope.
  4. Substitute the values of ( x_1 ), ( y_1 ), and ( m ) to find the equation of the tangent line.

By performing these steps, you'll obtain the equation of the tangent line at the given point.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7