What is the equation of the tangent line of #r=cot^2theta - sintheta# at #theta=pi/4#?

Answer 1

#y~~-4.707x+4.707pi/4+0.293#

#(dr)/(d theta)=-2cot(theta)csc^2(theta)-cos(theta)=r'(theta)#
#r'(pi/4)=-2/tan(pi/4)*1/sin^2(pi/4)-cos(pi/4)#
#=-4-sqrt(2)/2~~-4.707#
This is our instantaneous rate of change (slope of the tangent) when #theta=pi/4#

Then, we can just a point-slope formula to figure out the equation of that tangent line by plugging in values into the original function:

#y-y_1=m(x-x_1)#
#r(pi/4)=cot^2(pi/4)-sin(pi/4)=1-sqrt(2)/2~~0.293#

Thus:

#y-0.293=3.293(x-pi/4)#
#y~~-4.707x+4.707pi/4+0.293#
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 to the curve ( r = \cot^2(\theta) - \sin(\theta) ) at the point ( \theta = \frac{\pi}{4} ), we first find the slope of the tangent line by taking the derivative of ( r ) with respect to ( \theta ) and then evaluate it at ( \theta = \frac{\pi}{4} ). Then we use the point-slope form of the equation of a line to find the equation of the tangent line.

Taking the derivative of ( r ) with respect to ( \theta ) yields:

[ \frac{dr}{d\theta} = \frac{d}{d\theta}(\cot^2(\theta) - \sin(\theta)) ]

[ = \frac{d}{d\theta}(\cot^2(\theta)) - \frac{d}{d\theta}(\sin(\theta)) ]

[ = -2\cot(\theta)\csc^2(\theta) - \cos(\theta) ]

Evaluating this derivative at ( \theta = \frac{\pi}{4} ) gives:

[ \left. \frac{dr}{d\theta} \right|_{\theta=\frac{\pi}{4}} = -2\cot\left(\frac{\pi}{4}\right)\csc^2\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) ]

[ = -2(1)\left(\frac{\sqrt{2}}{2}\right)^2 - \frac{\sqrt{2}}{2} ]

[ = -1 - \frac{\sqrt{2}}{2} ]

Now, we have the slope of the tangent line. To find the equation of the tangent line, we use the point-slope form of a line with the point ( (\theta, r) = \left(\frac{\pi}{4}, \cot^2\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right)\right) ) and the slope we just found.

Substituting the values into the point-slope form, we get:

[ r - \left(\cot^2\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right)\right) = \left(-1 - \frac{\sqrt{2}}{2}\right)(\theta - \frac{\pi}{4}) ]

[ r - \left(1 - \frac{\sqrt{2}}{2}\right) = \left(-1 - \frac{\sqrt{2}}{2}\right)\left(\theta - \frac{\pi}{4}\right) ]

[ r = \left(-1 - \frac{\sqrt{2}}{2}\right)\left(\theta - \frac{\pi}{4}\right) + 1 - \frac{\sqrt{2}}{2} ]

This is the equation of the tangent line to the curve ( r = \cot^2(\theta) - \sin(\theta) ) at the point ( \theta = \frac{\pi}{4} ).

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