Angle between two curves (1 Viewer)

[Forbidden.]

One of the most disturbing things I find when checking my answers is when my responses are only close to the answers in the back of a text book.

Our current topic, before the exams is the "Angle Between Two Curves"


Formula for the angle between two curves

tanθ = |(m₁-m₂)/(1+(m₁m₂)|

Could someone verify my answers and check who is correct ?


Q4.

Find the acute angle between the curves y=x³ and y=x²-2x+2 at their point of interesction


My response:

x³=x²-2x+2

x³-x²+2x-2=0

x²(x-1)+2(x-1)=0

(x-1)(x²+2)=0

x-1=0 or x²+2=0

x=1 or x²=-2

However, square of numbers cannot be nagative

x=1

y=(1)³=1
y=(1)²-2(1)+2=1

d/dx (x³)
d/dx (x²-2x+2)


dy/dx (3x²)
dy/dx (2x-2)

At (1,1)

m₁=3
m₂=0

tanθ = |(m₁-m₂)/(1+(m₁m₂)|
tanθ = |(3-0)/(1+(3)(0))|
tanθ = 3
tanθ^-1 = 79⁰31'

My final answer: θ = 79⁰31'
Textbook's answer: θ = 71⁰34'


Q5.

What is the obtuse angle between the curves ƒ(x)=x²-4x and g(x)=x²-x+4 at the point where they meet ?

My response:

x²-4x=x²-x+4
-4x=-12
x=3

ƒ(x) (x²-4)
g(x) (x²-12)

ƒ'(x) (3x²) = 2(3)-4
m₁=2

g'(x) (2x-2) = 2(3)
m₂=6

tanθ = |(m₁-m₂)/(1+(m₁m₂)|
tanθ = |(2-6)/(1+(2)(6))|
tanθ = 4/13
tanθ^-1 = 19⁰

Let α be the obtuse angle

180-θ=α

α = 160⁰59'31"
α = 161⁰ (to the nearest degree)


My final answer: α = 161⁰
Textbook's answer: α = 162⁰

EDIT: Note my answers are in Gradient, not Degrees, my mistake everyone. I do hope my working out is of high quality

 

[Forbidden.]

For Q4, your working is fine up until the very last line. Just try calculating tan^-1(3) again. According to the calculator in my PC, it is 71.565... degrees (i.e., the same as the answer in the textbook.)

My calculator display:

tan^-1(3):

79.51672353

Pressing the ⁰' " button gives...

79⁰31'0.2"

And it's in ...

OMG ... It's in Gradient,

I swear, I wil further restrict lending my calculator to someone from now on ..

Now that I change it to Degrees ...

71.56505118

71⁰33'54.18"


Thanks ...

 

[Forbidden.]

How many times has that happened to me...

My teacher was telling me to find the limit of (sinx)/x as x -> 0 and I kept getting a weird answer. I was like WTF?! Until I realised I didn't have it on radians mode.

Shit happens, but now I reviewed my working out, it is all correct ..

 

[Riviet]

You should try and get into the habit of checking what mode your calculator is on whenever you're about to perform any trig operations.