# slope

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• Apr 10th 2007, 12:35 PM
jeph
slope
i have to find the slope of the tangent line at theta = pi/2 of the graph r=2 + 2cos(theta).

i did the x=rcos(Theta) and y=rsin(theta) thing then the dx/d(theta) dy/d(theta) thing, then dy/dx

i ended up with 4cos^2(theta) + 2cos(theta) - 2 for dy and dx = -2sin(theta) - 4cos(theta)sin(theta).

the slope is -2/-2 = 1 but how do you calculate 4cos^2(pi/2)?
• Apr 10th 2007, 12:54 PM
qbkr21
Re:
• Apr 10th 2007, 12:58 PM
qbkr21
Re:
Ohh...I didn't finish it up that pi/2 comes out. If it makes it any easier for you you and take pi/2 and multiple it by 180/pi and turn it into radians.
• Apr 10th 2007, 01:11 PM
jeph
so does the ^2 just change into a 2 or something and you multiply the 4 by that to get 8?
• Apr 10th 2007, 01:21 PM
qbkr21
Re:
Quote:

Originally Posted by jeph
i have to find the slope of the tangent line at theta = pi/2 of the graph r=2 + 2cos(theta).

i did the x=rcos(Theta) and y=rsin(theta) thing then the dx/d(theta) dy/d(theta) thing, then dy/dx

i ended up with 4cos^2(theta) + 2cos(theta) - 2 for dy and dx = -2sin(theta) - 4cos(theta)sin(theta).

the slope is -2/-2 = 1 but how do you calculate 4cos^2(pi/2)?

Yes. You leave constants alone. If you were to differentiate 4cos^2(pi/2), you would just be using the chain rule to for cos^2(pi/2). The 2 comes out in front of the constant, 4, that you left alone and makes an 8. Then you rewrite the 8 cos^2(pi/2) then differentiate cos^x(pi/2). Again you would rewrite the problem for the 3rd time and differentiate with respect to pi/2. Then rewrite the problem for the 4th time, thus giving you your final answer.
• Apr 10th 2007, 01:47 PM
Jhevon
Quote:

Originally Posted by qbkr21
I Jeph!!

Here is the work:

Hi qbkr21 and jeph!

qbkr21, usually you would be correct, but here i think we are dealing with polar coordinates, so things will be different.

am i right jeph, is this question in polar coordinates? if so, i will type up a solution to the problem
• Apr 10th 2007, 01:53 PM
qbkr21
Re:
What do mean what would it be like with Polar? I have never used them before...
• Apr 10th 2007, 01:59 PM
jeph
im getting confused.....is there a formula for calculating cos^2 problems? this differentiating stuff is gonna kill me on the exam :(
• Apr 10th 2007, 02:01 PM
qbkr21
Re:
Relax Jeph... You will do well just remember:

cos^2(x)= (cos(x))^2

Always Always Always now can you differentiate it?

:eek: WARNING cos^2(x) IS NOT EQUAL TO cos(x)^2:eek:
• Apr 10th 2007, 02:01 PM
Jhevon
Here are the graphs to give you an idea of how different the answers will be if we use the wrong coordinate system
• Apr 10th 2007, 02:04 PM
Jhevon
Polar coordinates is just a different type of coordinate system than the usual cartesian coordinates. I was introduced to it in calc 3, but i think my school starts to do it in calc 2 now. see Polar coordinate system

but don't worry, you should see it soon
• Apr 10th 2007, 02:05 PM
qbkr21
Re:
I guess polar coordinates are not functions since they do not appear to be one-to-one.
• Apr 10th 2007, 02:07 PM
Jhevon
Quote:

Originally Posted by qbkr21
I guess polar coordinates are not functions since they do not appear to be one-to-one.

correct. they always revolve. the r in a polar equation means radius, so it is based on circles and trig functions
• Apr 10th 2007, 03:42 PM
Jhevon
Quote:

Originally Posted by jeph
i have to find the slope of the tangent line at theta = pi/2 of the graph r=2 + 2cos(theta).

i did the x=rcos(Theta) and y=rsin(theta) thing then the dx/d(theta) dy/d(theta) thing, then dy/dx

i ended up with 4cos^2(theta) + 2cos(theta) - 2 for dy and dx = -2sin(theta) - 4cos(theta)sin(theta).

the slope is -2/-2 = 1 but how do you calculate 4cos^2(pi/2)?

just in case.

we can't really draw a line in polar coordinates, so we have to convert to rectangular coordinates before we can find the desired line.

r = 2 + 2cos(theta)
=> r^2 = 2r + 2rcos(theta) ............multiplied both sides by r

now going from polar to rectangular coord:

r^2 = x^2 + y^2

x = rcos(theta)

y = rsin(theta)

so r^2 = 2r + 2rcos(theta)
=> x^2 + y^2 = 2(x^2 + y^2)^(1/2) + 2x
now we differentiate implicitly, we get:

2x + 2y y' = (2x + 2y y')(x^2 + y^2)^(-1/2) + 2

=> 2x + 2y y' - (2x + 2y y')(x^2 + y^2)^(-1/2) = 2 .........factor out the (2x + 2y y'), we get
=> (2x + 2y y')(1 - (x^2 + y^2)^(-1/2)) = 2
=> 2x + 2y y' = 2/[1 - (x^2 + y^2)^(-1/2)]
=> 2y y' = 2/[1 - (x^2 + y^2)^(-1/2)] - 2x
=> y' = 2/{2y[1 - (x^2 + y^2)^(-1/2)]} - x/y

this will give us the slope of the tangent line. but what are the x and y values to plug in? let's find them.

we want theta = pi/2

if theta = pi/2 then r = 2 + 2cos(pi/2) = 2 + 0 = 2
so r = 2

so we have
x = 2cos(theta)
y = 2sin(theta)

so when theta = pi/2

x = 2cos(pi/2) = 0
y = 2sin(pi/2) = 2

so x = 0 and y = 2 at the point we are concerned with. so we get

y' = 2/{2(2)[1 - (0^2 + 2^2)^(-1/2)]} - 0/2
=> y' = 1 ..............the slope of the tangent line when theta = pi/2
• Apr 10th 2007, 06:08 PM
jeph
so would 4cos^2(theta) + 2cos(theta) -2 be equal to 4cos(theta)cos(theta) + 2cos(theta) -2? 0.o

oh i didnt see there was a second page. we learned polar coordinates in trig and now we are doing things with it in calc 2. i think its the differentiating thats getting me right now. i forget these things way too quickly after not doing it for a while...
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