Question

Help me finish a tangent line approximation problem! Will medal and fan!

Answer 1

Suppose the point ( (pi)/3 , (pi)/4 ) is on the curve sin x/x + sin y/y = C where C is a constant. Use the tangent line approximation to find the y-coordinate of the point on the curve with x-coordinate (pi)/3 + (pi)/180.

Answer 2

Okay, fist we must find the value of C, we have the function and a point on it, just plug these values and find C

Answer 3

So far, I know that the derivative is f'(x)= (y^2 (-xcosx + sinx))/ (x^2 (ycosy -siny) and I assume that I will use the formula f(a) + f'(a)(x-a)

Answer 4

Well, that is a possible solution, however, I know they are looking for me just to assume C is a constant and then use tangent line approximation rather than the exact answer that would give

Answer 5

Now, make the implicity differentiation on this function to find y'(x), then apply the tangent line aproximation that is:
y(x) = y(a) + y'(a)*(x-a)

Answer 6

so y in ((pi)/3 + (pi)/180, y) would be y(x)=Y(a)+ y'(a)*(x-a)?

Answer 7

if they gave you this point in the curve is for you to find the value of C

Answer 8

yes

Answer 9

so it would look like y(x)= (sin x/x + sin y/y) + (y^2 (-xcosx + sinx))/ (x^2 (ycosy -siny) * ((pi/3 + pi/180)-pi/3) with x= pi/3 and y = pi/4 ??

Answer 10

@M4thM1nd ?

Answer 11

the problem here is that we can't write the function y(x) = f(x) explicitly

Answer 12

so what would I have to do to make this correct?

Answer 13

I am a little confused with how to really go about this problem

Answer 14

i think we should write another funtion g(x,y) = sin(x)/x + sin(y)/y - C

Answer 15

i'm not sure

Answer 16

@SolomonZelman

Answer 17

@Marki

Answer 18

@hartnn

Answer 19

@SolomonZelman Is there any chance you could help walk me through this problem?

Answer 20

\(\large\color{slate}{\displaystyle\frac{\sin x}{x} + \frac{\sin y}{y} = C}\)

the point ( (pi)/3 , (pi)/4 ) is on the curve

\(\large\color{slate}{\displaystyle\frac{3\sin (\pi/3)}{\pi} + \frac{4\sin (\pi/4)}{\pi} = C}\)

\(\large\color{slate}{\displaystyle\frac{3}{\pi\sqrt{2}} + \frac{4\sqrt{3}}{2\pi} = C}\)

\(\large\color{slate}{\displaystyle\frac{3\sqrt{2}}{2\pi} + \frac{4\sqrt{3}}{2\pi} = C}\)

\(\large\color{slate}{\displaystyle\frac{3\sqrt{2}+4\sqrt{3}}{2\pi} = C}\)

So,

\(\large\color{slate}{\displaystyle\frac{\sin x}{x} + \frac{\sin y}{y} = \frac{3\sqrt{2}+4\sqrt{3}}{2\pi} }\)

this is your awesome function.
https://www.desmos.com/calculator/edhroqg1tl

when x= (pi)/3 + (pi)/180 = 61pi/180

Answer 21

this circle-like function obviously doesn't pass the vertical line test, and there can be two values with such an x coordinate, and those are 2 different tangent lines. So your question should be 2 point `s` with an x coordinate (pi)/3 + (pi)/180

Answer 22

This makes sense, but doesn't really fit the concept of tangent line approximation? I had found the derivative and solved it out for the slope of the tangent line which I plugged back into a y-y1=x(x-x1) equation to find the tangent line and then plugged in the x value (pi/3+pi/180) to find y... not very accurate considering the nature of this circle-type function... but, does it make sense to do it like this because I am on a tangent line approximation chapter?

Answer 23

wait, your dy/dx was ?

Answer 24

f'(x)= (y^2 (-xcosx + sinx))/ (x^2 (ycosy -siny)

Answer 25

Also, the function you solved for does not have a point (pi/3, pi/4)?

Answer 26

\(\large\color{black}{ \displaystyle \frac{ y\cos(y)- \frac{ dy}{dx}\sin(y)}{y^2}+\frac{ x\cos(x)- \sin(x)}{x^2}=0}\)

\(\large\color{black}{ \displaystyle \frac{ dy}{dx}\frac{ -\sin(y)}{y^2}+\frac{ y\cos(y)}{y^2}+\frac{ x\cos(x)- \sin(x)}{x^2}=0}\)

and solve for dy/dx, I got to go to the movie with my family right now, sorry