Help me solve this, im stuck

[Sophie12aronian]

 

[Deleted member 4993]

View attachment 21774

Is your given expression:

[f(xy)]⁴ *y³ - k*x = cos(y⁵ - y³)​

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this problem.

 

[Sophie12aronian]

My expression is [f(xy)]: 4y3 - k*x = cos(y5 - y3)

And well I'm not very good at math so I have problems understanding how to find the equation step by step

 

[BraveHeart258]

plug in y=0.5 and k=4.61 into the equation and break it down such that you isolate f(x,y) as a variable function. such that f(x,y)=equation. simplify, solve, plug, and chug.

 

[Sophie12aronian]

Y= -1.51598x +0.33702

Would this be correct?

 

[Deleted member 4993]

My expression is [f(xy)]: 4y3 - k*x = cos(y5 - y3)

And well I'm not very good at math so I have problems understanding how to find the equation step by step

you write:
[f(xy)]: 4y3 - k*x = cos(y5 - y3)

Are those supposed to be exponent? If yes, then you should express those as:

[f(xy)]: 4y^3 - k*x = cos(y^5 - y^3)

For computers ^ indicates exponent.

The first thing would be to understand that the normal line is perpendicular to the tangent. Next step would be to recall that slope of the tangent line at a given point P is same as the derivative of the curve at that point.

So if we are looking at a tangent at (x1, y1) for curve y = f(x) the equation of the tangent line would be:

(y - y₁) = m * (x - x₁) ....................................................... where m = [df/dx] at (x₁,y₁)

The equation to the normal line would be

(y - y₁) = (-1/m) * (x - x₁)

steps to solve would be:

• you will calculate x₁ from given y₁
  
  
• you will calculate [df/dx]
  
  
• calculate the equation for normal.

that's it .......

 

[Deleted member 4993]

Y= -1.51598x +0.33702

Would this be correct?

Please share your work for getting the equation?

 

[Sophie12aronian]

Ok so this is what I did

f(x,y): 4 y^3 - 4.61 x = cos (y^5-y^3)

solve for x

x = 0.867679 y^3 - 0.21692 cos(y^3 - y\^5)..........(*)

found the derivative wrt y

dx/dy = 2.60304 y^2+(0.650759 y^2-1.0846 y^4) \\sin (y^3-y^5)

value at y = 0.5

I got 0.659643

the normal is perpendicular to the tangent, thus the slope is

\- 1/0.659643 = -1.51597

x where y = 0.5

plugged y = 0.5 in the (*)

x = -0.107507

Used equation of a line passing trough (p,q) having slope m

y - q = m(x - p)

y - 0.5 = -1.51597(x - ( -0.107507))

y = -1.51598 x + 0.33702

 

[Deleted member 4993]

Ok so this is what I did

f(x,y): 4 y^3 - 4.61 x = cos (y^5-y^3)

solve for x

x = 0.867679 y^3 - 0.21692 cos(y^3 - y\^5)..........(*)

found the derivative wrt y

dx/dy = 2.60304 y^2+(0.650759 y^2-1.0846 y^4) \\sin (y^3-y^5)

value at y = 0.5

I got 0.659643

the normal is perpendicular to the tangent, thus the slope is

\- 1/0.659643 = -1.51597

x where y = 0.5

plugged y = 0.5 in the (*)

x = -0.107507

Used equation of a line passing trough (p,q) having slope m

y - q = m(x - p)

y - 0.5 = -1.51597(x - ( -0.107507))

y = -1.51598 x + 0.33702

I did not check the numbers - but the steps look correct to me.

 

[Sophie12aronian]

Tnahk you!!

 

[HallsofIvy]

My expression is [f(xy)]: 4y3 - k*x = cos(y5 - y3)

That's not much better! First, if you can't use "Latex" please indicate exponents by "^": 4y^3- kx= cos(y^5- y^3). Second, I have no idea what "[f(xy)]:" means. You said "curve" so you must not mean that f is a function of x and y which would give a surface, not a curve. Also "y= 0.5" is NOT a "point". a point needs both an x and a y.

I suspect there is NO "f" at all, that y is an implicit function of x and it is the derivative of y that you want to find.

And well I'm not very good at math so I have problems understanding how to find the equation step by step

 

[BraveHeart258]

I got \(\displaystyle = -8 - \dfrac {1}{4.61x} \)............................ WHAT is it equal to?

where

\(\displaystyle \cos(\dfrac{3}{24})=1\) ................................. INCORRECT

 

[BraveHeart258]

I would suggest never entering in values until you simplify. It makes it easier in physics as well.

 

[Deleted member 4993]

I got \(\displaystyle = -8 - \dfrac {1}{4.61x} \)

where

\(\displaystyle \cos(\dfrac{3}{24})=1\)

What is :

\(\displaystyle = -8 - \dfrac {1}{4.61x} \) - WHAT is it "equal to"

Then you write

\(\displaystyle \cos(\dfrac{3}{24})=1\).............................That is INCORRECT

 

[BraveHeart258]

I performed the calculation I received cos(3/24) ≈ 1 ≈ 0.999997
Do you disagree?

I can say that using significant figures cos(3/24)=1

 

[Deleted member 4993]

I performed the calculation I received cos(3/24) ≈ 1 ≈ 0.999997
Do you disagree?

I can say that using significant figures cos(3/24)=1

Where did you get that number.

First of all that should be cos(-3/24)

And you did not respond to:

I got \(\displaystyle =−8−14.61x=−8−14.61x\displaystyle = -8 - \dfrac {1}{4.61x} \)............................

WHAT is it equal to? What did you get - where did you get?

 

[Steven G]

I performed the calculation I received cos(3/24) ≈ 1 ≈ 0.999997
Do you disagree?

I can say that using significant figures cos(3/24)=1

Might you be using degrees instead of radians like you should.
Please start trying to help. You have not made one correct post so far.

 

[Deleted member 4993]

Might you be using degrees instead of radians like you should.
Please start trying to help. You have not made one correct post so far.

That is what s/he did (I am sure), but in this case numerically does not matter in a significant way.

cos(1/8 radian) = cos(7.16197^o) = 0.992198

Another "mistake" was dropping negative sign - again does not matter numerically. However, I think the numbers were "blindly punched" in without "mental review".