© 1996 by Karl Hahn
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1a) Remember that you use the chain rule to find the derivative of a composite of two functions, that is to find the derivative of f(g(x)) when you know the derivatives of both f and g. So you have to identify what two functions h(x) is the composite of. In this example, h(x) indicates that you first add 1 to x, then square the result. So take g(x) to be adding 1 to x, and take f(x) as squaring x. Then clearly
h(x) = f( g(x) )We know the derivatives of both f and g.
f(g) = g2 f'(g) = 2g g(x) = x + 1 g'(x) = 1If you had a problem with those derivatives, you ought to review the material (either here or in your text book or both) on elementary derivatives.
The chain rule says to substitute g(x) in for x into f'(x), then multiply by g'(x). Taking the first step we have:
f'( g(x) ) = 2(x + 1)Then adding the second step to this, that is multiplying this result by g'(x) (which is always equal to 1 regardless of x), we get the same thing because we are simply multiplying by 1.
h'(x) = 2(x + 1)(1) = 2(x + 1)And that is the answer.
1b) Once again, most of the battle is won here once you identify what two functions h(x) is the composite of. We can see that in this problem, h(x) tells us to take 1 less than twice x, then apply the polynomial, x2 - 3x + 2 to the result. So let
f(g) = g2 - 3g + 2and let
g(x) = 2x - 1We know how to find the derivatives of these:
f'(g) = 2g - 3and
g'(x) = 2Now simply apply the chain rule to these expressions. That is substitute g(x) in for x into the expression for f'(g), then multiply the result by g'(x).
h'(x) = f'( 2x - 1 ) g'(x) = ( 2(2x - 1) - 3)(2) = 8x - 10
2) Same as the last two problems, only this time one of the functions that forms the composite is something you have not yet learned to take the derivative of. So I told you what that derivative is. Let
f(g) = sin(g)and let
g(x) = x2Then
f'(g) = cos(g)(which is the derivative I told you in the problem) and
g'(x) = 2xNow again, apply the chain rule by your g(x) in for g into the expression for f'(g) and then multiplying the result by g'(x).
h'(x) = f'( x2 ) g'(x) = cos( x2 )(2x)And that is the answer.
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Solution to Coached Exercise: Derivative of x1/n
Step 1: We have
g(x) = x1/n eq. 4.4-8aFor its inverse we have
f(g) = gn eq. 4.4-8band for the derivative of its inverse we have
f'(g) = ngn-1 eq. 4.4-8c
Step 2: Take the composite of the two functions using the f and g symbols. If we put the inverse function, f on the outside, we have
f( g(x) ) = ??Do you remember that if you take a function of x and then take the inverse function of the result, you always get back the original x? So we have
f( g(x) ) = x eq. 4.4-9
Step 3: We take the derivative of both sides of 4.4-9. On the left side we apply the chain rule. That means we take the derivative of the outside function and multiply it by the derivative of the inside function. So we have left side of the equal
f'( g(x) ) g'(x)On the right side of the equal, we have the derivative of x, which we know is always 1. So we have
f'( g(x) ) g'(x) = 1 eq. 4.4-10
Step 4: Substitute back into 4.4-10 from 4.4-8a and 4.4-8c. Starting with substituting in 4.4-8c, which is an expression for f'(x)
n gn-1(x) g'(x) = 1 eq. 4.4-11aAnd the substituting back 4.4-8a for g(x), we get
n (x1/n)n-1 g'(x) = 1 eq. 4.4-11bDo you recall from algebra what it is you do when you encounter one power raised to another? You multiply the exponents. So we can simplify 4.4-11c to
n x(n-1)/n g'(x) = 1 eq. 4.4-11c
Step 5: Solve for g'(x). That's easy. Simply divide both sides of 4.4-11c by all the stuff on the left that isn't g'(x). You get
1
g'(x) = eq. 4.4-12a
n x(n-1)/n
and that is a legitimate answer. But most instructors will want you to
recall a little more algebra. Remember that one over a power is the same
as minus that power. That is
1
= a-n
an
So you can use that identity to simplify 4.4-12a to
1
g'(x) = x(1-n)/n eq. 4.4-12b
n
which is the more traditional answer.
Observe that (1-n)/n is exactly 1 less than our original exponent, 1/n. Observe also that the multiplier that is out in front of the x term in 4.4-12b is exactly equal to our original exponent. Remember that the rule for taking the derivative of x to a power was to subtract 1 from the power and multiply the resulting expression by the original power. That is the derivative of xn is nxn-1. And prior to this exercise we had proved it for n being both positive and negative integers. Now it seems that we have proved that the same rule applies when we allow n to be 1 over any integer (except zero).
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Solution to 2nd Coached Exercise: derivative of xm/n
where m and n are both integers.
Step 1: What are the two functions whose composite is xm/n? Recalling from algebra that this expression is the same as
(x1/n)mwe can see that this is a composite of:
g(x) = x1/n eq. 4.4-14awhich is the inner function, and
f(g) = gm eq. 4.4-14bwhich is the outer function.
Step 2: Find the derivatives of 4.4-14a and 4.4-14b. In the previous exercise we found the derivative of x1/n
1
g'(x) = x(1-n)/n eq. 4.4-15a
n
The other derivative we have seen many times before
f'(g) = mgm-1 eq. 4.4-15b
Step 3: Writing the composite function, h(x). This is easy.
h(x) = f( g(x) ) eq. 4.4-16
Step 4: Apply the chain rule to find h'(x). You should be getting good at this part by now.
h'(x) = f'( g(x) ) g'(x) eq 4.4-17
Step 5: Substitute back. I suggested you do it in stages, so we'll do it that way. We'll start out by substituting f'(x) in from 4.4-15b
h'(x) = m gm-1(x) g'(x) eq. 4.4-18aThen substitute g(x) from 4.4-14a
h'(x) = m (x1/n)m g'(x) eq. 4.4-18bThen substitute g'(x) from 4.4-15a
1
h'(x) = m (x1/n)m-1 x(1-n)/n eq. 4.4-18c
n
Step 6: Use algebra to simplify 4.4-18c. First remember about raising a power to a power -- you multiply the powers:
1
h'(x) = m x(m-1)/n x(1-n)/n eq. 4.4-19a
n
You can combine the multiplier, m, with the multiplier, 1/n,
to get
m
h'(x) = x(m-1)/n x(1-n)/n eq. 4.4-19b
n
And finally, recall from algebra that if you multiply two power expression
(that have the same base) you simply add the exponents. The two exponents
here are already over a common denominator, so this is straightforward.
m
h'(x) = x(m-n)/n eq. 4.4-19c
n
which is the answer that would be graded correct on an exam.
Can you determine for yourself that we have now proved that the rule for taking the derivative of xn (that is, the derivative of it is nxn-1) applies even when n is the quotient of any two integers (except for when the divisor is zero)?
Here is an even bigger challenge for you. We found the derivative by observing that
xm/n = (x1/n)mBut if you recall from algebra, it is also true for a power of a power that it doesn't matter which order you apply them in. This is an exception to the rule that you cannot generally reverse the order of applying functions when taking a composite (that is, f( g(x) ) is not generally equal to g( f(x) )). So it is true that
xm/n = (xm)1/nCan you start from this and derive 4.4-19c? Try it.
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Answers to More Chain Rule Exercises
3a) Observe that this is a composite of square root on the outside and 1 - x2 on the inside. Let f(x) = √x and let g(x) = 1 - x2. Using formulas that we have encountered so far, we know that
1
f'(g) =
2√g
and g'(x) = -2x.
Taking the composite of f and g have
h(x) = f( g(x) )When we apply the chain rule we see that
h'(x) = f'( g(x) ) g'(x)Substituting back using the expressions above for f'(g), g'(x) and g(x), we have
1
h'(x) = (-2x)
2√1 - x2
which is a legitimate answer, but some instructors would want you to simplify
to
-x
h'(x) =
√1 - x2
3b) Again we can see that this is a composite with square root on the outside and a polynomial on the inside. Let f(x) = √g and let g(x) = x3 - 7x2 + 3x - 4. Again we can use formulas that we know to find both f'(g) and g'(x).
1
f'(g) =
2√g
and g'(x) = 3x2 - 14x + 3. Now form the
composite
h(x) = f( g(x) )and then apply the chain rule to get
h'(x) = f'( g(x) ) g'(x)Substitute back your expressions for f'(x), g'(x), and g(x).
1
h'(x) = (3x2 - 14x + 3)
2√x3 - 7x2 + 3x - 4
which is a satisfactory answer.
3c) As in problem 3a, the inside function is g(x) = 1 - x2. The outside function is f(g) = g3/2. We know that g'(x) = -2x. From the discussion we have had in this section, we know that the formula of the derivative of xn being nxn-1 will work even when n = 3/2 (or when n is any rational number). That means that f'(g) = (3/2) g1/2 = (3/2) √g.
We have h(x) = f( g(x) ). So we apply the chain rule to this and find that
h'(x) = f'( g(x) ) g'(x)Now substitute f', g', and g back in to get
______
h'(x) = (3/2)√1 - x2 (-2x)
which is a legitimate answer, but your instructor will probably want you to
do the cancellation to get
______
h'(x) = -3√1 - x2 (x)
4) If you got this one, it means that you understand well what I have explained to you so far about the chain rule. I started you off by telling you to let f(g) = sin(g) and let g(x) = sin-1(x). The derivative of f(g) is given in the problem: f'(g) = cos(g). We don't know what g'(x) is yet -- that is what the problem is asking for.
I said that you should form the composite of f and g. Since f and g are inverses of each other, if follows that their composite must be equal to x. So we have
f( g(x) ) = xWe now apply the chain rule to find the derivative of the left side of this equation, and we know that on the right, the derivative of x is always 1. Hence
f'( g(x) ) g'(x) = 1Now substitute back. First substitute for g(x)
f'( sin-1(x) ) g'(x) = 1Now substitute f'(x)
cos( sin-1(x) ) g'(x) = 1Now here is the trick that unties the knot. Remember I reminded you that sin2(x) + cos2(x) = 1 for all x. If you apply a little algebra to that you can finagle it into
___________
cos(x) = √1 - sin2(x)
So we
substitute that identity in for cos
____________________
√1 - sin2( sin-1(x) ) g'(x) = 1
But what is sin2( sin-1(x) )?
It is the square of a composite of sin and its inverse. We know
that the composite of anything with its inverse is always equal
to x. So the square of that is simply x2.
When we apply that simplification we get
______
√1 - x2 g'(x) = 1
Now all we have left to do is solve for g'(x), which is easy.
1
g'(x) =
√1 - x2
And that is your answer. Strange, isn't it, that an inverse trig function
can have a derivative that is a simple algebraic function. This is your
first clue as to how calculus ties together all the common functions,
including algebraic functions, trig functions, log and exponential functions,
into one great family.
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Answers to Yet More Exercises
5a) Let g(x) = 3x + 4 and let f(x) = x3. Applying our usual formulas for derivatives, we have g'(x) = 3 and f'(x) = 3x2. Observe that h(x) is the composite of f and g, so
h(x) = f( g(x) )Now apply the chain rule to get
h'(x) = f'( g(x) ) g'(x)Substitute back the expressions we have for f', g', and g, and we get
h'(x) = 3(3x + 4)2 (3) = 9(3x + 4)2
5b) Again let g(x) = 3x + 4 but let
_
f(g) = √g
Applying our usual formulas for derivatives,
we have g'(x) = 3 and
1
f'(g) =
2√g
Observe that h(x) is the composite of
f and g, so
h(x) = f( g(x) )Now apply the chain rule to get
h'(x) = f'( g(x) ) g'(x)Substitute back the expressions we have for f', g', and g, and we get
3
h'(x) =
2√3x + 4
5c) This time let g(x) = x3 + 4x and again let
_
f(g) = √g
Applying our usual formulas for derivatives,
we have g'(x) = 3x2 + 4 and
1
f'(g) =
2√g
Observe that h(x) is the composite of
f and g, so
h(x) = f( g(x) )Now apply the chain rule to get
h'(x) = f'( g(x) ) g'(x)Substitute back the expressions we have for f', g', and g, and we get
3x2 + 4
h'(x) =
2√x3 + 4x
5d) This time let g(x) = x2 + 2x + 1 and let f(g) = g3/2. Applying our usual formulas for derivatives, we have g'(x) = 2x + 2 and
_
f'(g) = (3/2)√g
Observe that h(x) is the composite of
f and g, so
h(x) = f( g(x) )Now apply the chain rule to get
h'(x) = f'( g(x) ) g'(x)Substitute back the expressions we have for f', g', and g, and we get
___________
h'(x) = (3/2)√x2 + 2x + 1 (2x + 2)
The first way: If you multiply it out, then inside the outer parentheses you get -4x2 - 4x. So let g(x) = -4x2 - 4x and again let f(g) = g3/2. Applying our usual formulas for derivatives, we have g'(x) = -8x - 4 and
_
f'(g) = (3/2)√g
Observe that h(x) is the composite of
f and g, so
h(x) = f( g(x) )Now apply the chain rule to get
h'(x) = f'( g(x) ) g'(x)Substitute back the expressions we have for f', g', and g, and we get
3 _________
h'(x) = √-4x2 - 4x (-8x - 4)
2
The second way: We have a nested composite. Let u(x) = 2x + 1. Let g(u) = 1 - u2. And again let f(g) = g3/2. We have u'(x) = 2. We have g'(u) = -2u. And we have
_
f'(x) = (3/2)√x
Observe that h(x) is a nested composite of f, g,
and u. So
h(x) = f( g( u(x) ) )Applying the chain rule to the outside composite, we get
h'(x) = f'( g( u(x) ) ) × derivative of g( u(x) )To find the derivative of g( u(x) ) we apply the chain rule again to get g'( u(x) ) u'(x). So putting that all together we have
h'(x) = f'( g( u(x) ) ) g'( u(x) ) u'(x)Substituting back the expression u(x) and u'(x) we have
h'(x) = f'( g( 2x + 1 ) ) g'( 2x + 1 ) (2)Now substitute back expressions for f', g', and g we have
3 _____________
h'(x) = √1 - (2x + 1)2 (-2)(2x + 1) (2)
2
or, multiplying through all the 2's near the right, we have
3 _____________
h'(x) = √1 - (2x + 1)2 (-8x - 4)
2
I'll let you confirm that this answer is equivalent to the one we got
the first way.
-1
g'(u) =
2√u
And we have f'(g) = (2/3)g-1/3.
Observe that h(x) is the nested composite of
f, g, and u(x), so
h(x) = f( g( u(x) ) )Now apply the chain rule to the outside to get
h'(x) = f'( g( u(x) ) ) × derivative of g( u(x) )To find the derivative of g( u(x) ) we apply the chain rule again to get
h'(x) = f'( g( u(x) ) ) g'( u(x) ) u'(x)Substitute back the expressions we have for u' and u, and we get
h'(x) = f'( g( x2 + 1 ) ) g'( x2 + 1 ) (2x)Substituting further with expressions for f', g', and g, we have
______ -1
h'(x) = (2/3) ( a2 - √x2 + 1 )-1/3 (2x)
2√x2 + 1
6) The problem tells you that these two functions are inverses of each other, so when you take their composite, what do you get? The composite of inverse functions of x is always x. So let f(x) = ex and let g(x) = ln(x). The problem tells you that the derivative of ex is itself, so that gives us f'(x) = ex. We don't yet know what g'(x) is -- that is what the problem is asking for. We have
f( g(x) ) = xjust as we did in the other problems of this type that we did earlier. We use the chain rule to take the derivative of the left side, and, as in the other problems, the derivative of x is always 1. So we have
f'( g(x) ) g'(x) = 1Now substitute back the expressions we have for f' and g to get
eln(x) g'(x) = 1Notice that on the left we have a composite of the inverse functions, ex and ln(x). We know that that composite must always be equal to x. Hence the above equation becomes
x g'(x) = 1Solving for g'(x) is easy now
1
g'(x) =
x
And that is the derivative of ln(x).
7a) Here y is a function of x. So wherever we see an expression with y raised to a power, or some function of y, we will have to apply the chain rule. Here the only term like that is y3. That is a composite of f(y) = y3 and whatever function y(x) is. We know that f'(y) = 3y2. So we could attack y3 as
h(x) = f( y(x) )When we apply the chain rule to that we get
h'(x) = f'( y(x) ) y'(x)and substituting back we get
h'(x) = 3y2(x) y'(x)But we are using the notation that does away with the (x), so we have that the derivative of y3 is 3y2 y'. Putting that together with the derivatives of the other terms in this problem (which are expressions of the independent variable, x, only) we have
(3y2 y') + 2x = |
1 |
7b) In this one, it should be pretty clear that you will have to use the product rule. On the left of the equal we have the product of x and y, where y is a function of x. We know that the derivative of x is always 1. Since y is a function, all we can say about its derivative is that it is y'. Of course, to the right of the equal we have a constant, and the derivative of a constant is always zero. So applying the product rule to the left and zero as the derivative of the right we get
y + (x y') = 0
7c) You have to do a little more work to get this one. Again to the right of the equal we have a constant, and the derivative of a constant is always zero, so that is no problem. To the left we have a composite. Let g(x) = x2 + y2 (but I hear you asking, "Isn't g a function of both x and y?" Remember that y is a function of x, so whatever y(x) is, it can be written as some expression of x. And that's why, in the end, g is a function of x alone). Let f(g) = √g. The left hand side of this problem is clearly the composite of f( g(x) ), and so we apply the chain rule. We know that
1
f'(g) =
2√g
It's finding g'(x) that is the problem. But we can use the implicit
method on it, applying the chain rule
to the y2 term. Doing that gives
g'(x) = 2x + (2y y')Now we are in a position to apply the chain rule to f( g(x) ) and make the substitutions back:
1
f'( g(x) ) g'(x) = (2x + (2y y') ) = 0
2√x2 + y2
Notice that since the right hand side of the equal is zero, we can multiply
through by
_______
2√x2 + y2
and get the equivalent (and much simpler) relationship of
2x + (2y y') = 0
7d) This one I recommended you try two different ways. The first way was with the quotient rule, so here is that way. We have y as a function of x. The numerator of the left hand side is x + y, so the derivative of the numerator is 1 + y'. The denominator is 2x - y, so the derivative of the denominator is 2 - y'.
The quotient rule says, take the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all over the denominator squared. That will give us the derivative of the left hand side. The right hand side is a constant, so its derivative is zero. That gives us
( (2x - y) (1 + y') ) - ( (x + y) (2 - y') )Once again we have a zero on the right, so we can multiply through by (2x - y)2 to simplify this to= 0 (2x - y)2
( (2x - y) (1 + y') ) - ( (x + y) (2 - y') = 0We can multiply it out and gather like terms to further simplify
2x + 2xy' - y - yy' - 2x + xy' - 2y + yy' = 3xy' - 3y = 0Dividing out the 3 gives
xy' - y = 0which is about as simple as that expression can get. Now lets try it the other way. That is to multiply the original equation through by the denominator, 2x - y, and then take the derivative of both sides.
x + y = 2x - yTaking the derivative of both sides of that you get
1 + y' = 2 - y'"Wait a minute," you're saying. "Why are the two expressions different?" But the real question is, can they both be true?
What happens if we solve x + y = 2x - y for y? We get
1
y = x
2
That means that
1
y' =
2
for all x. Try substituting those expressions into both of our
answers, and you'll see that they work in both cases.
This reiterates what I said before, that doing implicit differentiation doesn't tell you all the information about the derivative (in this case we had to solve for y to get all the information), it just tells you something about it. And if you algebraically munge the equation before you do the implicit differentiation, the result you get might tell you something different about the derivative, but just as true.
7e) Looks like the product rule for both sides of this one. Note that both u and v are functions of the independent variable, t. That means that taking the derivative of any expression with either a u or v in it will cause a u' or v' respectively to pop out.
Using the chain rule and the rule that the derivative of xn is nxn-1 whenever n is a rational number, we can determine the derivatives of u1/2 and v1/3. They will be (1/2)u-1/2 u' and (1/3)v-2/3 v' respectively.
On the right hand side, we have that the derivative of u + v is u' + v' and that the derivative of t2 is 2t. Putting that all together using the product rule on both sides we get
( (1/2)u-1/2 u' v1/3) + (u1/2 (1/3)v-2/3 v') =
( (u' + v') t2) + ( (u + v) 2t )
7f) Again both u and v are functions of the independent variable, t. This one is all chain rule. On the left, let g(t) = u-1/2 + v-1/3, and let f(g) = g2. We know that f'(g) = 2g. Since both u and v are functions of t, taking the derivative of any expression that contains them will cause a u' or v' to pop out. So we get g'(t) = ( (-1/2)u-3/2 u') + ( (-1/3)v-4/3 v'). On the right hand side we have that the derivative of t2 is 2t. Applying the chain rule gives
f'( g(t) ) g'(t) = 2tSubstituting back with the expressions for f', g', and g gives
2 (u-1/2 + v-1/3) ( ( (-1/2)u-3/2 u') + ( (-1/3)v-4/3 v') ) = 2t
8) Egad! A word problem. Don't panic. Look carefully for the information the problem gives you. You have volume as a function of height: v(h) = (1/2)h2. The problem tells you that the cross sectional area (which is also a function of height, h, so we shall call it a(h)) is the derivative of that. So
a(h) = v'(h) = hWe know that the cross section is always a circle. From classes you took years ago you know that radius, r(h), follows a fixed relationship to area, a(h). That relationship is
a(h) = π r2(h)So, can you solve for r(h) from this? That's easy.
r(h) = |
r(h) = |
9) I assume you read the hint perhaps several times. We shall call the distance between the ships, s. It is a function of time, so every time you see s, you can imagine it to mean s(t). The derivative of s is the rate of change of distance between the two ships. We denote it to be s', but it too is a function of time, so every time you see s' you can imagine it to mean s'(t). In addition, xA, yA, xB, yB xA', xB', yA', and yB' are all functions of time as well, so you can imagine xA to mean xA(t), and likewise with all the others.
The hint says to use the Pythagorean formula for distance. You do recall it from algebra, don't you? So we have
s2 = (xB - xA)2 + (yB - yA)2 PythagorasNow you could take the square root of both sides of the above equation to get a solution for s, but the problem is simpler if you keep it like this for now. But do keep in mind that the square root of that nasty looking thing to the right of the equals is the solution for s.
Now take the derivative of both sides of the above equation. Use the chain rule and implicit differentiation to do it, remembering that all the variables shown are functions of time. So each variable has its own derivative. So, for example, the derivative of s2 using implicit differentiation is 2s s'.
The two summands on the right are each composite functions, that is they are each the square of the difference between two variables. And each of the variables is a function of time, so each has its own derivative. Hence, the derivative of (xB - xA)2 is 2(xB - xA)(xB' - xA').
It is important you see why this is. Applying the chain rule as we learned it here, we have the composite, f(g(t)), where f(t) = t2 and g(t) = xB(t) - xA(t). The derivative above is just an application of the chain rule to this composite.
So for the overall equation, taking the derivative of both sides of the Pythagorean distance formula, we have
2s s' = 2(xB - xA)(xB' - xA') + 2(yB - yA)(yB' - yA') deriv. of PythYou can immediately cancel the 2's to get
s s' = (xB - xA)(xB' - xA') + (yB - yA)(yB' - yA') deriv. of PythNow let's use this plus what we know to answer part a. If you set Ship A up at the origin, then at t = 0 the problem gives:
- xA = 0
- yA = 0
- xA' = 15 knots
- yA' = 0
- xB = 0
- yB = 20 nautical miles
- xB' = 0
- yB' = -25 knots
Now start substituting. We get
s s' = (0 - 0)(0 - 15 knots) + (20 mn - 0)(-25 knots - 0)Of course on the left we have s s' and what we are really looking for is s. But s is just the distance the two ships are apart. And the problem says that right now (interpret that to mean at t = 0) Ship A is 20 nautical miles south of Ship B. So s(0) = 20 nm. Dropping the stuff that's multiplied by zero and substituting 20 nm for s and dividing it out of both sides, we get
s'(0) = -25 knotsAnd that is the answer to part a. The ships are closing at 25 knots.
To solve part b, you have to use what you know about rate problems. The list of values above gives you the position coordinates of each ship at time of "right now," which means at t = 0. Here is how you compute them for other times:
- xA(t) = xA(0) + t xA'
- yA(t) = yA(0) + t yA'
- xB(t) = xB(0) + t xB'
- yB(t) = yB(0) + t yB'
- xA(t) = 0 + (t × 15 knots)
- yA(t) = 0 + (t × 0 knots)
- xB(t) = 0 + (t × 0 knots)
- yB(t) = 20 - (t × 25 knots)
s s' = (xB - xA)(xB' - xA') + (yB - yA)(yB' - yA') deriv. of Pythand solve for the t that makes s' = 0.
And what about that pesky s on the left hand side? It is only zero if the ships are colliding. And if it's not zero, we don't care what it is. If s' = 0, that forces the right hand side to be zero also. So we just solve for the right hand side being equal to zero. Here goes:
0 = (0 - t×15 knots)(0 - 15 knots) + (20 - t×25 knots - 0)(-25 knots - 0)
0 = 225t - 500 + 625t
500 = 850t
10
t = hours = 0.588 hours
17
After all this discussion, part c is a piece of cake, eh? Just
look at
s s' = (xB - xA)(xB' - xA') + (yB - yA)(yB' - yA') deriv. of Pythone more time. And remember, if the ships aren't colliding, then s can't be zero. So once more, for s' to be zero, the right hand side must be zero also. And the right hand side is exactly the expression given in part c.
And because we never took the square root of both sides of the Pythagorean formula, we never had to take the square root of anything. That turned out to be a shortcut.
I know these exercises haven't been very colorful or exciting. Consider what we have done so far on derivatives to be the calisthenics you need to do to get in shape for the good stuff. It is important that taking derivatives becomes second nature for you. Memorize the rules and learn to recognize at a glance where each rule applies. Much of what will follow will depend upon this skill. If you are not keen with it, you will be muddling with taking the derivative of something, and that will occupy so much of your mind that you will miss some other concept that that derivative was illustrating. You and your study partners should be drilling each other on taking derivatives by now.
In the next section, we will see one of the ways we can apply derivatives to problems in the real world. It ought to be less drudgery than this section was. See you there.
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