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-BARSIC- [3]
3 years ago
13

The slope of the tangent line to the curve x^3y+y^2-x^2=5 at the point (2,1) is

Mathematics
1 answer:
zheka24 [161]3 years ago
5 0

Answer:

-4/5

Step-by-step explanation:

To find the slope of the tangent to the equation at any point we must differentiate the equation.

x^3y+y^2-x^2=5

3x^2y+x^3y'+2yy'-2x=0

Gather terms with y' on one side and terms without on opposing side.

x^3y'+2yy'=2x-3x^2y

Factor left side

y'(x^3+2y)=2x-3x^2y

Divide both sides by (x^3+2y)

y'=(2x-3x^2y)/(x^3+2y)

y' is the slope any tangent to the given equation at point (x,y).

Plug in (2,1):

y'=(2(2)-3(2)^2(1))/((2)^3+2(1))

Simplify:

y'=(4-12)/(8+2)

y'=-8/10

y'=-4/5

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The system is:

i)    <span>2x – 3y – 2z = 4
ii)    </span><span>x + 3y + 2z = –7
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the last equation can be simplified, by dividing by -2, 

thus we have:

</span>i)    2x – 3y – 2z = 4
ii)    x + 3y + 2z = –7
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The procedure to solve the system is as follows:

first use any pairs of 2 equations (for example i and ii, i and iii) and equalize them by using one of the variables:

i)    2x – 3y – 2z = 4   
iii)   2x +2y +z = -5 

2x can be written as 3y+2z+4 from the first equation, and -2y-z-5 from the third equation.

Equalize:  

3y+2z+4=-2y-z-5, group common terms:
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similarly, using i and ii, eliminate x:

i)    2x – 3y – 2z = 4
ii)    x + 3y + 2z = –7

multiply the second equation by 2:


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Let's use elimination method, multiply the equation a by -2:

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Solution: (x, y, z) = (-1, 0, -3 ) 


Remark: it is always a good attitude to check the answer, because often calculations mistakes can be made:

check by substituting x=-1, y=0, z=-3 in each of the 3 equations and see that for these numbers the equalities hold.</span>
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