Equation of tangent line to sin(xy) + y = sqrt(3)/2 + pi/3

[clmason]

I am having trouble with this question.

1) Let the curve be defined as sin(xy) + y = sqrt(3)/2 + pi/3. Find an equation of the tangent line to C at point (1,2).

I know that I need to take the derivative and evaluate for the given point. What I am a little confused on is..do I need to set the equation to zero and then take the derivative, or do I solve for y and then go from there?

Any help is appreciated.
Thanks,
-CM

 

[stapel]

1) Let the curve be defined as sin(xy) + y = sqrt(3)/2 + pi/3. Find an equation of the tangent line to C at point (1,2).

From your questions, it sounds as though you missed the lectures covering the chapter on implicit differentiation...?

It would take much too long to attempt to cover that material here, so please study at least three online lessons. Once you're familiar with the process, the following will make sense:

Differentiate implicitly with respect to x. Solve for "dy/dx =". This is, as always, your slope expression.

You are given the point of tangency, and thus the values of x and y. Plug these in to find the numerical value of the slope.

Then follow the customary process for finding the equation of a straight line, given a point and a slope. :wink:

 

[Deleted member 4993]

I am having trouble with this question.

1) Let the curve be defined as sin(xy) + y = sqrt(3)/2 + pi/3. Find an equation of the tangent line to C at point (1,2).

I know that I need to take the derivative and evaluate for the given point. What I am a little confused on is..

do I need to set the equation to zero and then take the derivative, <<< Absolutely not - cannot be justified under any circumstances

or

do I solve for y and then go from there? <<< You could try - but I think that solving for 'y' would be very hard if not impossible for this problem.

Best way to do it would be differentiate it implicitly.

Any help is appreciated.
Thanks,
-CM

 

[BigGlenntheHeavy]

sin(2)+2 does not equal sqrt(3)/2+PI/3.

 

[Deleted member 4993]

sin(2)+2 does not equal sqrt(3)/2+PI/3.

That result tells you that (1,2) is not on the graph. I "assume" there is a typo - it meant the tangent is drawn "from" the point (1,2).

 

[galactus]

The right side is a constant will go to 0.

\(\displaystyle sin(xy)+y\)

Chain rule on sin(xy):

\(\displaystyle cos(xy)(\overbrace{xy'+y}^{\text{product rule on xy}})=xy'cos(xy)+ycos(xy)\)

The derivative of y is y'.

Put it together:

\(\displaystyle xy'cos(xy)+ycos(xy)+y'=0\)

Now, solve for y'

 

[Deleted member 4993]

Clmason,

Can you please check the problem for correctness?

Since (1,2) is not on the graph of function, this can turn into a bear of a problem.

 

[BigGlenntheHeavy]

\(\displaystyle sin(xy)+y = sqrt(3)/2+Pi/3\)
\(\displaystyle y = 1.4255x+.5745 (about)\)
\(\displaystyle y = 2, x = 1\)