What is the slope of the polar curve #f(theta) = theta^2-theta - cos^3theta + tan^2theta# at #theta = pi/3#?

Question

Emma Aldridge · Accepted Answer

The slope, #m ~~ 2.7#

Slope of a tangent to a polar curve

From the reference we have the equation:

#dy/dx = ((dr(theta))/(d theta)sin(theta) + r(theta)cos(theta))/((dr(theta))/(d theta)cos(theta) - r(theta)sin(theta))#

We are given #r(theta)#:

#r(theta) = theta^2 - theta - cos^3(theta) + tan^2(theta) #

Use WolframAlpha to compute #(dr(theta))/(d theta)#:

#(dr(theta))/(d theta) = 2 θ+3 sin(θ) cos^2(θ)+2 tan(θ) sec^2(θ)-1#

Use WolframAlpha to evaluate #(dr(pi/3))/(d theta) ~~ 15.6#

Evaluate #r(pi/3) ~~ 2.9#

When we evaluate the remaining parts of #dy/dx# at #pi/3#, we obtain the slope, m, of the tangent line.

#m = (15.6sin(pi/3) + 2.9cos(pi/3))/(15.6cos(pi/3) - 2.9sin(pi/3))#

#m ~~ 2.7#

Charles Arocha · Answer

undefined To find the slope of the polar curve at a specific point, you can first compute the derivative of the polar function with respect to theta and then evaluate it at the given theta value.

The derivative of the polar function f(theta) = theta^2 - theta - cos^3(theta) + tan^2(theta) with respect to theta is given by:

f'(theta) = 2theta - 1 + 3cos^2(theta)sin(theta) + 2tan(theta)sec^2(theta)

Now, you can plug in theta = pi/3 into this derivative expression and calculate the value to find the slope of the curve at theta = pi/3.