We want to find the smallest positive value
for theta for which the tangent line to the curve
r equals two times e raised to the power of 0.3 theta
is horizontal.
Well the polar curve is graphed here below,
and notice how it appears that the angle theta
we're looking for
would be approximately here
in the second quadrant where we'd have
a horizontal tangent line at this point.
Remember if we have a horizontal tangent line,
that means dy dx would be equal to zero.
Because we have a polar curve,
dy dx is equal to dy d theta
divided by dx d theta which gives us this quotient here.
Let's review where this comes from.
Assume we have r equals f of theta,
which means r is a function of theta.
We know for polar coordinates x equals r cosine theta
and y equals y sine theta.
But if r equals f of theta, we can write x as
x equals f of theta times cosine theta,
and we can write y equals f of theta times sine theta.
And therefore to find dy d theta,
we'd have to apply the product rule
which gives us this numerator here.
We also have to apply the product rule
to find dx d theta using our equation here for x
which gives us our denominator.
So going back to our example,
notice here we have r equals f of theta
which is equal to two times e
raised to the power of 0.3 theta
and therefore f prime of theta
is going to be equal to two times e
raised to the power of 0.3 theta
times the derivative of 0.3 theta with respect to theta,
that'd be 0.3.
And therefore f prime of theta
is equal to two times 0.3 would be 0.6
e to the power of 0.3 theta.
Now we have all the information we need to find dy dx,
so we'll find our derivative,
set it equal to zero and solve for theta.
So we have dy dx is equal to
the numerator's going to be f prime of theta
times sine theta which would be 0.6 times
e to the 0.3 theta times sine theta.
And then we have plus f of theta,
so plus two e to the 0.3 theta times cosine theta
divided by f prime of theta times cosine theta
which would be 0.6 e to the 0.3 theta cosine theta,
and then we have minus f of theta times sine theta
which would be two e to the 0.3 theta sine theta.
And again, because we're trying to find
where we have a horizontal tangent line,
we set this equal to zero and solve for theta.
Let's go ahead and do this on the next slide.
So to solve this equation, we know a fraction
is only going to be zero when the numerator is zero
and the denominator is nonzero.
So in order to solve this equation here,
we can set the numerator equal to zero and solve for theta.
If you're still not convinced,
we could write zero in fraction form,
for example, zero over one, cross multiply,
and solve the resulting equation.
But notice how we'd have the numerator times one
is equal to the denominator times zero,
and therefore we're just solving the equation
where the numerator is equal to zero.
Notice how the exponential part is a common factor.
Let's go ahead and factor that out.
So we have e to the 0.3 theta
times the quantity 0.6 sine theta
plus two cosine theta equals zero.
Well, this product equals zero
when either factor is equal to zero.
But this exponential factor is never going to be zero,
and therefore to solve this equation,
we just need to set this factor here equal to zero
and solve for theta.
So we have 0.6
sine theta
plus two cosine theta equals zero.
Let's go ahead and subtract two cosine theta
on both sides of the equation,
that would give us 0.6 sine theta
is equal to negative two cosine theta.
Now from here remember sine theta divided by cosine theta
is equal to tangent theta.
So notice how if we divide both sides by cosine theta.
Here we have tangent theta.
But let's also divide by a factor of 0.6 on both sides.
So now if we simplify,
notice the left side here simplifies to tangent theta
and the right side simplifies just negative two
divided by 0.6.
Let's finish this on the next slide.
If tangent theta is equal to negative two
divided by 0.6, we can take the inverse tangent
or arctangent on both sides of the equation
and therefore theta is going to be equal to
inverse tangent or arctangent
of negative two divided by 0.6.
And now from here, we'll have to go to the calculator
to get a decimal approximation.
But remember inverse tangent only return angle
in the first or fourth quadrant.
And we're looking for an angle in the second quadrant.
Let's verify we are in radian mode.
So we'll press the MODE key, notice radian is highlighted,
so we'll go back to the home screen,
we'll press 2ND Tangent, negative two
divided by .6, closed parenthesis and ENTER.
So if we round to four decimal places,
this would be approximately negative 1.2793.
Notice this angle is terminal in the fourth quadrant,
not the second quadrant.
If we go back to our graph,
the calculator gave us the angle in the fourth quadrant
somewhere down here where we do have
a horizontal tangent line,
but the angle the calculator gave us
is not the smallest positive value.
So in order to find this angle here in the second quadrant,
we'll have to take the angle the calculator gave us
and add pi radians, which will give us the angle here
in this second quadrant.
So going back to the calculator,
since we already have this value in here,
we can just press plus and then 2ND caret
or 2ND exponent key here for the pi, press ENTER.
This is the angle we're looking for in the second quadrant.
It's approximately 1.8623.
This would be the smallest positive value for theta
where we have a horizontal tangent line.
So our solution is theta is approximately 1.8623 radians.
I hope you found this helpful.