C Roettger, S15
Math 165, Project 2A – solutions
Details only for project 2A – for the others, see the table below.
Project 2A A spaceship is has coordinates (x, y) satisfying the equation
x2 − xy + y 4 = 91
a) Suppose that there is a function y = f (x) which makes this equation true
for all values of x in an interval. Find a formula for f 0 (x) using implicit
differentiation.
b) Find an equation for the tangent line to the curve given above at the point
p0 = (5, 3).
c) Use the equation from part b) to approximate the spaceship’s position at
a time when x = 5.002.
d) Sketch or plot the curve given above, at least a part of it close to (5, 3).
You can either solve the equation for several values of x to get some points on
the curve, or use a computer package like SAGE, MATLAB, or Mathematica.
e) What can you say about the interval of x-values where your formula for
f 0 (x) from part a) makes sense?
Solution for Project 2A. a) Use implicit differentation. This means
differentiate both sides with respect to x to get
2x − (xy 0 + y) + 4y 3 y 0 = 0
and solve for y 0 to get
y0 =
y − 2x
.
4y 3 − x
b) Use part a) and substitute x = 5, y = 3 to get
y 0 (5) = −
7
.
103
Then the tangent line at x = 5 can be described by the equation
y=−
7
(x − 5) + 3.
103
7
c) y(5.002) ≈ − 103
(5.002 − 5) + 3 ≈ 2.9986.
For d), see the plot below. For e), look at the plot.
Project 2A – Path of spaceship with tangent line at the point (5, 3)
e) There are no points on the curve with x outside an interval which is
about (−10, 10). Inside this interval, the upper branch of the distorted circle
path forms a perfectly sensible graph of a function. Note how this chunk
of the curve has vertical tangent lines at the endpoints, so points where the
denominator 4y 3 − x of y 0 (x) equals zero. Coupled with the equation for the
curve, this would be a tool for computing the points with vertical tangents
more precisely.
Solutions for Projects 2B, C, D
The table below gives the solutions to a) and b) for projects B, C, D. Plug
in the given x-value into the tangent line equation to get c).
Project a)
B
C
D
b)
85y
−
12x
231
y0 =
y=
(x − 2) + 3
3
24y − 85x
478
9y − 6x2
15
y0 =
y = (x − 2) + 1
4
10y − 9x
8
2
43y
−
18x
76
y0 =
y = − (x − 3) + 2
3
24y − 43x
63
Project 2B – Path of spaceship with tangent line at the point (2, 3)
e) The path of the spaceship near (2, 3) would be the graph of a differentiable
function defined up to about x = 65 (at the end, there is a vertical tangent
line). It could be extended down to x = 0, ending with another vertical
tangent line. But the function could actually be extended to the left down to
about x = −65, choosing the vertical one of the two branches of the curves
meeting at (0, 0) – it √
would still be continuous but not differentiable at the
origin, similar to y = 3 x.
Project 2C – Path of spaceship with tangent line at the point (2, 1)
e) The path of the spaceship near (2, 1) would be the graph of a differentiable
function defined up to about x = 2.2 (at the end, there is a vertical tangent
line). It can be extended forever to the left, choosing the horizontal one of
the two branches of the curves meeting at (0, 0).
Project 2D – Path of spaceship with tangent line at the point (3, 2)
e) The path of the spaceship near (3, 2) would be the graph of a differentiable
function defined up to about x = 3.1 (at the end, there is a vertical tangent
line). It could be extended down to x = 0, ending with another vertical
tangent line. But the function could actually be extended forever to the
left, choosing the lower of the two branches of the curves meeting at (0, 0)
– it would
still be continuous but not differentiable at the origin, similar to
√
3
y = x.