Find eguation of tangent to each curve at a given point:

2x^2y^3 + 3xy+5=0 (1,-1)

The only thing that throws me off is two variables

y=2sin(x) + cos^2x (pi/6, 7/4)

I am pretty sure I know how to do this, just want to double check.

Thx in advance

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- Apr 12th 2009, 04:43 PMBobCalcFind eguation of tangent
Find eguation of tangent to each curve at a given point:

2x^2y^3 + 3xy+5=0 (1,-1)

The only thing that throws me off is two variables

y=2sin(x) + cos^2x (pi/6, 7/4)

I am pretty sure I know how to do this, just want to double check.

Thx in advance - Apr 12th 2009, 04:48 PMmr fantastic
- Apr 12th 2009, 04:58 PMBobCalc
Mr F says: Use implicit differentiation to get dy/dx. Then proceed in the usual way.

2x^2y^3 + 3xy+5

Ok here is the dy/dx 12xy^3+ 3

I just do not know what to do because there are 2 variables x and y, should I isolate for y?

y=2sin(x) + cos^2x (pi/6, 7/4)

dy/dx= 2cox - 2sinxcox

After plugging in pi/6, I get root 3- root 3/2- This is the only thing I think I can be wrong at.

And thanks for the fast response mr.F - Apr 12th 2009, 05:07 PMmr fantastic
- Apr 13th 2009, 07:53 AMBobCalc
Hopefully this differentiation is correct. Math is not my strong point in school, so sorry for making mistakes.

4xy^3 + 6y^2x^2 +3y+3x=0

I sub in In (1,-1)

-4+6+3-3=0

**2**

So m is 2? - Apr 13th 2009, 12:08 PMBobCalc
I really need someone to check my differentiation, my teacher told me that similar question is going to be on the test, so I really need to know.

Find eguation of tangent to each curve at a given point:

2x^2y^3 + 3xy+5=0 (1,-1)

4xy^3 + 6y^2x^2 +3y+3x=0

I sub in In (1,-1)

-4+6+3-3=0

**2**

So m is 2? Any help would be greatly**appreciated** - Apr 13th 2009, 05:57 PMmr fantastic
I asked if you had been taught implicit differentiation. If you do not understand how to apply that technique, you are not going to be able to do these questions. Go back to your class notes and/or textbook and review all the examples you have been given. Here is one more example:

Differentiate both sides with respect to x, treating y as an implicit function of x.

The derivative of is .

In the above I have used the product rule to differentiate and I have used the chain rule to differentiate . Note: .

The derivative of is . I used the product rule here.

So you end up with: .

Substitute x = 1 and y = -1: .

Solve for . That is your value of m. - Apr 13th 2009, 06:17 PMBobCalc
I am very sorry, I though you asked if I knew just differentiation. This is my first year taking calculus and I

**never learned implicit differentiation**.Its not in my textbook either, because I am done with calc and there are no questions like that.

Do I just go -4-1 -3-3

Do I just sub in -1/1 for dy/dx

Once again my bad - Apr 13th 2009, 06:38 PMmr fantastic