Math 126 Homework hints
Note: As I’m sure you’ve all seen in section, I make mistakes too! I can’t guarantee that everything
here will be right 100% of the time. If you think there’s an error in something, please e-mail me
and I’ll try to get a correction up. Same goes for any clarification or further questions; if I’ve made
something even more confusing, please e-mail or come to office hours to clear it up!
14.3 Part II
Problem 8: Find the indicated partial derivative.
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f (x, y, z) = exyz ; fxyz
Recall that in this notation, we read the variables with respect to which we are taking partial
derivatives from left to right. So, this expression is saying “first differentiate with respect to x,
then y, then z. So, what’s fx ?. If we treat yz 4 like a constant, then the chain rule just pulls that
constant out in front when we differentiate with respect to x:
fx = yz 4 exyz
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Now things get a little bit trickier. This time, we have to use the product rule. Think of this as
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the product of two functions in y, so fx = (yz 4 )exyz . Treating x and z as constants now, if we
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differentiate yz 4 then all that’s left is the z 4 term. On the other hand, differentiating exyz with
respect to y gives us a new coefficient of xz 4 , again by the chain rule. So, the product rules tells
us:
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fxy = z 4 exyz + (yz 4 )(xz 4 )exyz
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= z 4 exyz + xyz 8 exyz
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I’ll finish up by just doing the derivative of the first term with respect to z. This is almost exact
like how we just found the derivative with respect to y. We need to use the product rule on the
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product of the two functions z 4 and exyz . The derivative of the first one is just 4x3 . For the
second one, we again use the chain rule and treat x and y as constants. When we do this, what’s
the derivative of xyz 4 with respect to z? It’s just 4xyz 3 . Putting this together, the derivative of
the first term above is just:
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(ez 4 )exyz + (z 4 )(4xyz 3 exyz )
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Now only the last term xyz 8 exyz is left. The process will be very similar to that of the first
term.
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Problem 11: The wind-chill index is modeled by the function
W = 13 + 0.6T − 11v 0.1 + 0.4T v 0.1
where T is the temperature (C) and v is the wind speed (km/h). When T = 8C and v = 14 km/h, by
how much would you expect the apparent temperature W to drop if the actual temperature decreases
by 1C? What if the wind speed increases by 1 km/h?
Ok, the wording on this one is kind of weird. The first thing to do is figure out exact what it’s
asking. This “apparent temperature” function is a function of both the temperature T and the
wind speed v. The first question is asking how much W changes if T goes down by one. So, first
we need to find a function that tells us the change in W with respect to T . Sound familiar? Let’s
calculate the partial derivative of W with respect to T :
∂W
= .6 + .4v .1
∂T
Now what’s this rate of change when T = 8 and v = 14? If actually doesn’t matter what T is, as
you can see above. Plug in v = 14:
∂W
= .6 + .4(14.1 )
∂T
This tells us the apparent temperature W increases by this amount for every degree that T increases.
So, this is also the amount that W decreases for every degree that T decreases.
For the second part, we’re trying to find how much W changes if v changes. For this, calculate
the derivative of W with respect to v, ∂W
∂v . One word of warning: pay attention to the power rule
when you differentiate the v 0.1 term. Plug in your values T = 8 and v = 14, and you’re good to
go.
Problem 12: The paraboloid z = 5 − x − x2 − 2y 2 intersects the plane x = 3 in a parabola. Find
parametric equations in terms of t for the tangent line to this parabola at the point (3, 2, 15). (Enter
your answer as a comma-separated list of equations. Let x, y, and z be in terms of t.)
Calculating a tangent line to some parabola isn’t that bad, but the first thing we need to do is find
out what the equation of this parabola even is. We need to find all the point on the paraboloid
where x = 3. So, just plug in x = 3 in the equation for the paraboloid and see what kind of
equation we get:
z = 5 − (3) − (32 ) − 2y 2
z = −7 − 2y 2
The slope of the tangent line is the derivative of this function, so
dz
= −4y
dy
We want the line at (3, 2, 15), so y = 2 and the slope of the tangent line is −8 at this point. Now
we’ve got all the information that we need. Use the point-slope formula:
(z − (−15)) = −8(y − 2), so z = −8y + 1
All that’s left is to parametrize the line with respect to t. First off, we have x constant at x = 3.
The choices for y and z are a little bit arbitrary; we could set either one of them equal to t and
solve for the other to get parametrizations. Since we already have a nice formula for z in terms of
y above, let’s let y = t. Now what’s z in terms of t?
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14.4 Part I
Problem 9: Use differentials to estimate the amount of metal in a closed cylindrical can that is
30 cm high and 4 cm in diameter if the metal in the top and the bottom is 0.1 cm thick and the
metal in the sides is 0.05 cm thick. (Round your answer to two decimal places.)
Think about the formula for the volume of a cylinder: V = πr2 h. What’s the differential dV ? First
take the derivative with respect to r and multiply by dr, then with respect to h and multiply by
dh:
dV = 2πrh dr + πr2 dh
This is fine, but what exactly is this? If we have some cylinder of radius r and height h and we
increase the radius by an amount dr, then the first term is roughly telling us how much volume
we’ve added. Likewise, if we increase the height by some amount dh, then this is telling us roughly
how much the volume has increased. So, think about starting with our cylinder that is 30 cm high
and 4 cm in diameter. This has radius 2, so:
dV = 2π(30)(2) dr + π(22 ) dh
dV = 120π dr + 4π dh
Now think of measuring the amount of metal on the top as being the increase in volume if we were
to add the thickness 0.1 to the height, and measuring the amount of metal on the sides as being
the increase in volume if we were to add the thickness .05 to the radius...
Note: If you think back to Math 125, this formula 2πrh dr is similar to how you derived the method
of cylindrical shells. In that case, it looked like 2πxf (x) dx (we were viewing one shell as a cylinder
of radius x, height f (x), and thickness dx).
Problem 10: Suppose you need to know an equation of the tangent plane to a surface S at the
point P (4, 1, 4). You don’t have an equation for S but you know that the curves
r1 (t) = h4 + 2t, 1 − t2 , 4 − 5t + t2 i
r2 (u) = h3 + u2 , 2u3 − 1, 2u + 2i
both lie on S. Find an equation of the tangent plane at P .
First, both r1 and r2 pass through this point P :
r1 (0) = h4, 1, 4i
r2 (1) = h3 + 1, 2 − 1, 2 + 2i
So, both of the tangent vectors to r1 and r2 at P are parallel to the tangent plane at this point.
Let’s calculate those vectors:
r01 (t) = h2, −2t, −5 + 2ti so r01 (0) = h2, 0, −5i
r02 (u) = h2u, 6u2 , 2i, so r02 (1) = h2, 6, 2i
Now this reduces to a problem from the first couple of weeks. We have to vectors that lie in a
plane, and we know a point on the plane. How can we get the equation from a plane out of this
information?
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