Question

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

Answer 1

`-3*sqrt(2)`

Explanation:

First, we need to find the derivative of the equation with respect to `theta`

`d/(d theta) r=12*d/(d theta)[sin(theta/3)*cos(theta/2)]`

where we have already factored out the constant. Using the chain rule we get two terms:

`d/(d theta) r=12*[d/(d theta)sin(theta/3)]*cos(theta/2)`

`+ 12* sin(theta/3)*[d/(d theta) cos(theta/2)]`

Each of the derivatives in the square brackets can be completed as follows:

`d/(d theta)sin(theta/3) = 1/3 cos(theta/3)`
and
`d/(d theta) cos(theta/2)=-1/2 sin(theta/2)`

plugging these back in above

`d/(d theta) r=4 cost(theta/3) cos(theta/2) - 6 sin(theta/3) sin(theta/2)`

next, we plug in the value of #theta for which we want the slope and simplify

`d/(d theta) r|_(theta = 3pi/2)= 4cos(pi/2)cos((3pi)/4) - 6sin(pi/2)sin((3pi)/4)`

`d/(d theta) r|_(theta = 3pi/2)= 0-6*sqrt(2)/2 = -3*sqrt(2)`