MATH 100 V1A
October 3rd – Practice problems
Solutions
1. The curve x2/3 +y 2/3 = 1 is called an astroid. Sketch the astroid and find the equations
of all lines of slope −1 that are tangent to the curve.
Solution: The astroid x2/3 + y 2/3 = 1 looks like:
2 y
1
x
−2
−1
1
2
−1
−2
Note that, based on our sketch of the astroid, we expect there to be two lines tangent
to the curve with slope −1.
Using implicit differentiation, we get
2 −1/3 2 −1/3 0
x
+ y
· y = 0,
3
3
so, solving for y 0 we find that
y0 = −
x−1/3
y 1/3
=
−
.
y −1/3
x1/3
Since we want to find the equations of tangent lines with slope −1, we set y 0 = −1
and find that x = y. Now, we need to make sure that the points where y 0 = −1 are
2/3
2/3
2/3
also on the astroid, so we substitute
x= y into
x + y = 1 to get that 2x = 1,
or x = ± 2√1 2 . So, the points 2√1 2 , 2√1 2 and − 2√1 2 , − 2√1 2 are the only points on the
curve where the slope of the tangent line is −1. The equations of the tangent lines are
given by
1
1
y − √ = −1 x − √
2 2
2 2
and
1
1
y + √ = −1 x + √ .
2 2
2 2
2. Explain why every curve in the set of curves x2 + y 2 = ax intersects every curve in the
set of curves x2 + y 2 = by at right angles. (Here a and b are any constants.)
Solution: First, if the curve x2 +y 2 = ax intersects the curve x2 +y 2 = by, then ax = by,
or y = ab x. This is a point on both curves if it also satisfies the equation of one of the
curves. That is, if
a 2
2
x = ax.
x +
b
2
Solving for x gives that either x = 0, or x = aa2 = a2ab+b2 . Since y = ab x, we get that
1+
these correspond to y = 0, or y =
a2 b
.
a2 +b2
b2
Now, two curves intersect at right angles if their tangent lines at the point of intersection meet at right angles. That is, if m1 is the slope of the first curve’s tangent line at
the point of intersection, and m2 is the slope of the second curve’s tangent line, then
we need to show that m1 · m2 = −1.
Using implicit differentiation, we get that for the first curve, x2 + y 2 = ax,
2x + 2yy 0 = a or y 0 =
a − 2x
,
2y
and for the second curve, x2 + y 2 = by,
2x + 2yy 0 = by 0
or y 0 = −
2x
.
2y − b
So, at the point (0, 0), the first curve has a vertical tangent line (“infinite” slope),
and the second curve has a horizontal tangent line (zero slope). So they meet at right
angles.
2
2
At the point a2ab+b2 , a2a+bb 2 , the first curve has tangent line with slope
0
a−2
y =
2·
ab2
a2 +b2
a2 b
a2 +b2
a(a2 + b2 ) − 2ab2
a2 − b 2
=
=
,
2a2 b
2ab
and the second curve has tangent line with slope
y =−
ab2
a2 +b2
a2 b
−
2
2
a +b
2·
0
2
b
=−
2ab2
2ab
=− 2
.
2
2
2
2a b − b(a + b )
a − b2
So, we see that the product of the two slopes is −1.
2
3. Show that on a circle of radius r centred at a point O, the tangent line at any point P
on the circle is perpendicular to the line joining O to P .
Solution: Let the point O have coordinates (x0 , y0 ), and the point P have coordinates
(x1 , y1 ). Then, the equation of the circle is given by
(x − x0 )2 + (y − y0 )2 = r2 .
Using implicit differentiation
2(x − x0 ) + 2(y − y0 ) · y 0 = 0 or y 0 = −
x − x0
.
y − y0
0
. Now, the slope of
So, the slope of the line tangent to the circle at P is m1 = − xy11 −x
−y0
y1 −y0
the line joining O to P is m2 = x1 −x0 . So, m1 · m2 = −1, and we can conclude that
the two lines are perpendicular.
4. A function f is said to have inverse function g if g(f (x)) = x for all x in the domain
of f and f (g(y)) = y for all y in the domain of g. Suppose f and g are differentiable.
Find a formula for g 0 (y) in terms of f 0 , g and y only (it should not contain x). (Hint:
Let y = f (x). What’s g(y)?)
Solution: Let y = f (x). Then g(y) = g(f (x)) = x, by definition of inverse function.
Differentiating both sides ofg(f (x)) = x using the Chain Rule, we get
g 0 (f (x)) · f 0 (x) = 1 or g 0 (f (x)) =
1
f 0 (x)
Now, if y = f (x), then g(y) = x (as above). So, we get that
g 0 (y) =
3
1
f 0 (g(y))
.
.