Question

Calculus 2 help.

Answer 1

So \(r\) and \(d\) are constant with respect to \(x\) and \(y\), right?

Answer 2

Yeah

Answer 3

\[
\frac{dy}{d\theta}=\frac{dy}{dx}\frac{dx}{d\theta}
\]

Answer 4

You can use this to solve for \(dy/dx\).

Answer 5

Any tangent line \(T\) is given by \[
T-y(x_0) = \frac{dy}{dx}(x_0)(x-x_0)
\]

Answer 6

so dy/dx=r+dsin(theetha)/r-dcos(theeta)

Answer 7

Didn't get your last point about the tangent line :/

Answer 8

The tangent line \(y\) of a function \(f(x)\) at the point \((x_0,f(x_0))\) is: \[
y-f(x_0)=f'(x_0)(x-x_0)
\]

Answer 9

Same concept, but different variables.

Answer 10

^ something new for me :/

Answer 11

It's calculus 1.

Answer 12

y-y1=m(x-x1) That? :/

Answer 13

Yes, but \(m=y'(x)\)

Answer 14

Yeah right.

Answer 15

What next?

Answer 16

First is finding \(dy/dx\)

Answer 17

You didn't get it right the first time.

Answer 18

This is wrong

dy/dx=r+dsin(theetha)/r-dcos(theeta)

Answer 19

Yes, that is wrong.

Answer 20

You said \(r\) and \(d\) are constants.

Answer 21

Confused :/

Answer 22

What is\[
\frac{dy}{d\theta}
\]

Answer 23

r+dSin(theetha)

Answer 24

No

Answer 25

That is wrong.

Answer 26

Thing about it.

Answer 27

You said \(r\) is a constant.
What is \[
\frac{dr}{d\theta }
\]

Answer 28

I don't know.

Answer 29

Derivative of a constant is \(0\).

Answer 30

But it with x na? :/

Answer 31

\[
y=r-d\cos \theta
\]

Answer 32

\[
\frac{dy}{d\theta}=d\sin \theta
\]

Answer 33

oh yeah :/

Answer 34

\[
x=r\theta - d\sin\theta
\]\[
\frac{dx}{d\theta }=r-d\cos \theta
\]

Answer 35

\[
\frac{dy}{dx}=\frac{d\sin\theta}{r-d\cos\theta }
\]

Answer 36

Okay now.

Answer 37

Okay never mind, it only wants the slope of the tangent line, not the tangent line itself. So we're good at \(dy/dx\).

Answer 38

Yes but what about b part.

Answer 39

To have a vertical tangent, then you'd need \[
r-d\cos\theta=0
\]Since the slope would be undefined at that point.

Answer 40

So no vertical tangent :)

Answer 41

Well, if \(d<r\) then \[
\forall \theta \quad d\cos\theta < r
\]So there is no way for it to reach \(0\).

Answer 42

A more formal proof\[\begin{split}
r &>d\geq d\cos\theta \\
r&> d\cos\theta\\
r- d\cos\theta &> 0

\end{split}\]