Question

for the curve x^3 + y^2 = x + y passing through (-2,3) use the tangent line at x = 2 to approximate the value of y at x = -1.5

Answer 1

implicit diff?

Answer 2

yes!

Answer 3

start with
\[3x^2+3y^2y'=1+y'\]

Answer 4

wait... where did you get the 3y'y'?

Answer 5

it is not \(3y'y'\) is it \(3y^2y'\)

Answer 6

right but why didn't you put 2yy' since its squared?

Answer 7

oh doh, because i read it wrong sorry
it is in fact what you said
\[3x^2+2yy'=1+y'\]

Answer 8

oh ok:)

Answer 9

so.... how do we approach it next?

Answer 10

i guess you need the slope of the line \(y'\) at the point \((-2,3)\)

Answer 11

now that i look at the problem i can't say it makes much sense
that slope will have nothing to do with the curve at \(x=2\)
are you sure it is not \(x=-2\)?

Answer 12

oh no you're right! it is -2! sorry!

Answer 13

whew
so now you have a choice
you can solve \[3x^2+2yy'=1+y'\] for \(y'\) and then put \(x=-2,y=3\) and see what you get

Answer 14

or you can put \(x=-2,y=3\) at the start and solve
\[3\times(-2)^2+2\times 3\times y'=1+y'\] for \(y\) using elementary algebra
either way will give \(y'\) at \((-2,3)\)

Answer 15

so is the answer y' = 11/12? and is that supposed to be the slope or did i completely miss this?

Answer 16

\[
12 + 6y' = 1 + y' \\
5y' = -11 \\
y' = -\frac{11}{5}
\]

Answer 17

oh jeez i multiplied wrong...

Answer 18

then find the equation of the line with slope
\[-\frac{11}{5}\] through the point \((-2,3)\)
leave it in the point slope form

Answer 19

\[
y' = \frac{y_2-y_1}{x_2-x_1} = \frac{y_2-3}{-1.5+2} = -\frac{11}{5}
\]

Answer 20

finally replace \(x\) by \(-1.5\)

Answer 21

\[(x_1, y_1) = (-2, 3) \\
(x_2,y_2) = (-1.5, y_2) \\
\text{Slope = }\frac{y_2-y_1}{x_2-x_1} = \frac{y_2-3}{-1.5+2} \]

Answer 22

-4.1 is y correct?

Answer 23

\[
\frac{y_2-3}{-1.5+2} = -\frac{11}{5} \\
y_2-3 = -2.2 * 0.5 = -1.1\\
y_2 = 3 - 1.1 = 1.9 \text{ or } \frac{19}{10}
\]

Answer 24

oh haha i subtracted the 3...

Answer 25

but thank you guys sooo much!!

Answer 26

you are welcome.