Question

y^6+x^3=y^2+10x ,normal at(0,1) at the given point,find the slope of the curve,the line that is tangent to the curve,or the line that is normal to the curve,as requested.

Answer 1

the derivative defines the slope ...

the equation of a line is from algebra classes so should already be known

a normal is a perp line, a line with a perpendicular slope

Answer 2

this is the problem i could not find it.

Answer 3

assuming that y is a function of x .... do an implicit

y^6+x^3=y^2+10x

6y^5 y' +3x^2 x' = 2y y' + 10x'

with respect to x, x'=dx/dx = 1

6y^5 y' +3x^2 = 2y y' + 10, we know (x=0, y=1)

6y'= 2y' + 10 , solve for y', to determine the slope at (0,1)

Answer 4

now you have a slope, and a point ( a y intercept at that) ... make the equation.

Answer 5

y=10/4,y=5/2

Answer 6

y-1=5/2(x-0)=5/2x+1

Answer 7

good, now what property do perp lines have? something about the slope ...

Answer 8

It is showing me answer supposed to be y=-2/5x+1

Answer 9

perp lines have negative reciprocal slopes, yes .. in other words, flip and negate

Answer 10

slope 5/2, for the tangent

slope -2/5, for the normal

Answer 11

It is very confusing to me.I need to surgery to shift my brain for this problems.:)

Answer 12

Thank you for your help.

Answer 13

:) since the problem is asking for the normal line at(0,1)

slope of the tangent is y'

slope of the normal is -1/y'

Answer 14

That' great,I need to explain everything how I did it,This is really help.

Answer 15

good luck

Answer 16

Tnx.