# Parametrics / Polars practice exam

*label*Mathematics

*timer*Asked: Dec 4th, 2017

*account_balance_wallet*$25

**Question description**

Do the problems in the attached study guide. Please show work. Can be handwritten or typed, I don't care.

## Tutor Answer

Complete.

150b Review

December 2017

Problem 1

a)

dy

=

dx

dy

dt

dx

dt

cos t

− sin t

= − cot t

=

At t =

π dy

4 , dx

= −cot

π

4

b)

From part a) we know the slope of the tangent line at t =

and y = 1 + √12 . Then solving for the y intercept,

1

1

1 + √ = (−1) √ + b

2

2

√

b=1+ 2

Then the tangent line is y = −x + (1 +

√

2)

c)

We proceed using the Pythagorean identity:

x2 = cos2 t

(y − 1)2 = sin2 t

x2 + (y − 1)2 = cos2 t + sin2 t = 1

Thus we have x2 + (y − 1)2 = 1

1

π

4 Att=π 4 ,

x=

√1

2

d)

The curve is a circle of radius 1 centered at (0, 1). We can recognize it from the

parametric equation since

Problem 2

The slope of the tangent line is given by the derivative.

dy

= 2t

dt

1 dy

dx

=

dt

t dx

=

dy

dt

dx

dt

= 2t2

dy

=2

At t = 1, (x, y) = (3, 3) so evaluating at t = 1, dx

Therefore, using the slope-intercept form, we have

3 = 2(3) − b =⇒ b = 3

So the tangent line is y = 2x − 3

Problem 3

a)

dy

= 4t3 − 10t

dt

dx

= 3t2 − 1

dt

dy

4t3 − 10t

=

dx

3t2 − 1

b)

The tangent lines are horizontal when the derivative is 0. This occurs when

4t3 − 10t = 0

2t(2t2 − 5) = 0

r

t = 0, ±

5

2

Substituting

q the equations for x and y, this yields (x, y) = (0, 4),

q

into

5

9

3

5

9

3

2

2, −4 , −2

2, −4 .

2

c)

The tangent lines are vertical when the denominator of the derivative is 0 since

it does not exist. Then 3t2 − 1 = 0 =⇒ t = ± √13 . Substituting into the

equations for x and y, we get (x, y) = ± √23 , 22

9

d)

Evaluating at t = 1, we have

dy

4(1)3 − 10(1)

=

= −3

dx

3(1)2 − 1

Problem 4

a)

dx

= r − r cos θ

dθ

dy

= r sin θ

dθ

dy

sin θ

=

dx

1 − cos θ

dy

At θ = π2 , dx

= 1 so the slope of the tangent line is 1.

At θ = π2 , (x, y) = πr

2 − r, r . Using the slope-intercept form, we have

πr

−r +b

r=

2

Therefore, b = r 2 − π2 so the tangent line is y = x + r 2 − π2 .

b)

Tangent lines are vertical when the derivative does not exist and horizontal when

the derivative is 0. Thus, sin θ = 0 =⇒ θ = nπ for n ∈ Z. However, 1 − cos θ =

0 has root θ = 2nπ. Therefore, we must check the limit using l’Hopital’s rule.

dy

θ

= cos

Taking the derivative of the numerator and denomoinator, dx

sin θ . At θ =

2nπ, the denominator goes to 0 so the derivative is undefined. Therefore, there

is a vertical tangent at θ = 2nπ. Substituting in for the coordinates, we have

vertical ta...

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