Math 126 Homework hints
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10.2 and 13.2
Problem 6: Find the equations of the tangents to the curve x = 6t2 + 4, y = 4t3 + 2 that pass
through the point (10, 6).
First of all, how do we find the tangent line to a point on a curve at time t? Recall our worksheet
from last week:
dy/dt
slope of tangent line =
dx/dt
In our case, dy/dt = 12t2 and dx/dt = 12t. Hence:
slope of tangent line =
12t2
=t
12t
At a time t, our y-coordinate is 4t3 + 2 and our x-coordinate is 6t2 + 4. So, using the point-slope
formula, the slope of the tangent line to the curve at time t is:
y − (4t3 + 2) = t(x − 6t2 − 4)
For what values of t will this line pass through the point (10, 6)? What will the equation be in
those cases?
Problem 7: Consider r(t) = h3t − 4, t2 + 4i. Finding r0 (t) for this problem is straightforward
calculation. As for identifying the picture of the graph, there’s one observation that should solve
it immediately: Which y values are possible for r(t)? Can you rule any of the pictures out based
on this information?
Problem 16: Evaluate the integral.
Z
8
8ti − t3 j + 5t9 k dt
0
Based on our integration of vector functions worksheet, we know that we can just split this up
into the component functions:
Z 2
Z 2
Z 2
3
9
i
8t dt − j
t dt + k
5t dt
0
0
0
1
Problem 19: If u(t) = hsin 6t, cos 6t, ti and v(t) = ht, cos 6t, sin 6ti, use the theorem to find
d
[u(t) × v(t)] .
dt
The formula in the theorem states:
d
[u(t) × v(t)] = u0 (t) × v(t) + u(t) × v0 (t)
dt
So, calculate u0 (t) = hcos 6t, − sin 6t, 1i and v0 (t) = h1, − sin 6t, cos 6ti. Find the appropriate cross
products, and add them together.
Problem 22: Find the distance traveled by a particle with position (x, y) as t varies in the given
time interval.
x = 2 sin2 t, y = 2 cos2 t, 0 ≤ t ≤ 3π
What is the length of the curve?
Recall the arc-length formula for curves (x(t), y(t)) from t = a to t = b:
Z bp
Length =
(x0 (t))2 + (y 0 (t))2 dt
a
Here, we calculate x0 (t) = 4 cos t sin t, y 0 (t) = −4 cos t sin t. Hence:
Z
3π
Length =
p
(4 cos t sin t)2 + (−4 cos t sin t)2 dt
0
Z
=
3π
p
32(cos t sin t)2 dt
0
This is most easily simplified with a trig identity: 2 cos t sin t = sin 2t, which tells us 4(cos t sin t)2 =
(sin 2t)2 . Hence:
Z 3π p
8(sin 2t)2 dt
0
2π
Z
√
2 2| sin 2t| dt
0
What does the graph of sin 2t look like? It repeats itself every π/2 radians, flipping sign each time.
By looking at this symmetry, this allows us to see that the integral above is the same as 6 times
the integral from 0 to π/2:
Z π/2 √
6
2 2 sin(2t) dt
0
This integral gives us the total distance traveled by the particle. What about the length of the
curve itself? The answer lies in the previous paragraph. We can see that the particle will repeat
its position every π/2 radians, so the curve itself is traced out only on this first interval...
2
10.3
Problem 6: Find a Cartesian equation for the curve and identify it:
r2 cos 2θ = 1
This problem is made easy with an appropriate trig identity. Using cos 2θ = cos2 θ − sin2 θ, we
get:
(r cos θ)2 − (r sin θ)2 = 1
Now transform this into x and y coordinates.
Problem 18: Find the points on the given curve where the tangent line is horizontal or vertical
for 0 ≤ θ < π:
r = 7 cos θ
We have an equation for the slope of the tangent line:
dy/dθ
dy
=
dx
dx/dθ
As in previous cases with functions x(t) and y(t), this tangent line will be zero if the numerator
is zero. So, what is dy/dθ? We know that y = r sin θ and that r = 7 cos θ by the equation of the
curve. Hence, y = 7 cos θ sin θ. Now differentiate and solve for θ (and then r) so that dy/dθ is zero
at (r, θ). Repeat for dx/dθ.
3