What is the slope of the tangent line of #r=-2sin(3theta)-12cos(theta/2)# at #theta=(-pi)/3#?

Answer 1

1) Differentiate both sides
2) Examine #(d)/(d theta) r((-pi)/3)#
3) Get that the slope is #3#

The slope of the tangent line of r at #theta = (-pi)/3#, is the same as the derivative of the function for that exact x-value. Therefore, we need to differentiate both sides of the function, so we get an expression for #d/(d theta)# r.
#d/(d theta) [r] = d/(d theta) [-2sin(3theta) - 12cos(theta/2)]# #d/(d theta) [r] = d/(d theta) [-2sin(3theta)] - d/(d theta)[-12cos(theta/2)]#
From this point, we need to know how to differentiate #sin theta# and #cos theta#:
#d/(d theta) sin theta = cos theta#
#d/(d theta) cos theta = -sin theta#
We need to know how the chain rule works as well. In this case, we have to substitute #3theta# with #u# and #theta/2# with #v#. When we then differentiate, we have to also differentiate our substitute. (If you want to have a further explanation, notify me).
We then have: #d/(d theta) [r] = d/(d theta) [-2sin(u)] - d/(d theta)[12cos(v)]# #d/(d theta) [r] = -2cos(u) * d/(d theta) [u] - (-12sin(v) * d/(d theta)[v])#
Let's now substitute #u# and #v# back to our original functions.
#d/(d theta) [r] = -2cos(3theta) * d/(d theta) [3theta] - (-12sin(theta/2) * d/(d theta)[theta/2])#
#d/(d theta) [r] = -2cos(3theta) * 3 + 12sin(theta/2) * 1/2 = -6cos(3theta) + 6sin(theta/2) #
Now we have an expression for # d/(d theta) r(theta)#, where we can put in any values we want, and get the slope for whatever #theta#-value we want. So let's put in #theta = (-pi)/3#:
#d/(d theta) [r((-pi)/3)]# = #d/(d theta) [r] = -6cos(3(-pi)/3) + 6sin((-pi)/(3*2))# #= -6cos(-pi) + 6sin(-pi/6)#
#cos (-pi) =-1# and #sin((-pi)/6) = -1/2#
This gives us that the slope, for #theta = (-pi/3)#, is: #-6(-1) + 6 * (-1/2) = 6-3 = 3#
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 slope of the tangent line of the polar curve r = -2sin(3θ) - 12cos(θ/2) at θ = -π/3, you would first find the derivative of r with respect to θ, then evaluate it at θ = -π/3 to get the slope of the tangent line at that 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 3

To find the slope of the tangent line of ( r = -2\sin(3\theta) - 12\cos(\frac{\theta}{2}) ) at ( \theta = -\frac{\pi}{3} ), you first need to find the derivative ( \frac{dr}{d\theta} ). Then evaluate the derivative at ( \theta = -\frac{\pi}{3} ) to get the slope of the tangent line at that 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