Query 2 revised

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course mth 174

4/22 11:44 pm

query 2#$&*

course mth 174

4/21 12:40 am

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Question: Problem 4 section 6.3.

6.3.8 (previously 6.3 #14, ds / dt = -32 t + 100, s = 50 when t = 0). Find the function s(t).

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Your solution:

We begin by finding the antiderivative of the function -32t+100

The antiderivative is -16t^2+100t+c which is equal to s(t)

We now have to substitute in for s being 50 and t being 0

50=-16(0)^2+100(0)+c

50=C

So finally s(t) = -16t^2+100t+50

confidence rating #$&*:

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Given Solution: 3

• s' = 100 - 32t.

Integrating with respect to t we obtain

• s= 100t - 16t^2 + C.

Since s = 50 when t = 0 we have

• 50 = 100(0) - 16(0)^2 + C,

which we easily solve to obtain

• 50 =C.

this into the expression for s(t) we have

• s(t) = 100(t) - 16t^2 + 50

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Self-critique (if necessary):

Ok except I wrote it as -16t^2+100t+50

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Self-critique Rating:

That's actually probably a better order. Either decreasing or increasing powers of t would be standard.

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Question: Problem 6 section 6.3

problem 6.3.17 was 6.3.6 (previously 6.3 #16) water balloon from 30 ft, v(t) = -32t+40

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Your solution:

We begin with -32t +40 we must find the general antiderivative

S(t)=-16t^2+40t+c

We are trying to find the velocity at 30ft and at time 0

S(0)=-16*0^2+40*0+c=30

C=30

So the antiderivative is -16t^2+40t+30

We need to now evaluat at 30 feet and time 0 by solving this equation

-16t^2+40t+30=0

This isn’t easily factorable by trial and error so we use the quadratic formula

-b+-squareroot (b^2+4ac)/2a

-40+-squareroot(1600-4*-16*30)/2*-16

-40+-squareroot of 3520/-32

-40+59.3/-32 or -40-59.3/-32

-.61 or 3.10 is the solution to this equation but now we must substitute both of these values into the original function

-32t + 40

-32(-.61)+40=59.52

-32(3.10)+40=-59.2

So the answer would be 59.52 ft per second.

confidence rating #$&*: 3

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Given Solution:

Since v(t) = s ' (t), it follows that the antiderivative of the v(t) function is the s(t) function so we haves

• s(t) = -16 t^2 + 40 t + c.

Since the building is 30 ft high we know that s(0) = 30.

Following the same method used in the preceding problem we get

• s(t) = - 16 t^2 + 40 t + 30.

The water balloon strikes the ground when s(t) = 0. This occurs when

-16 t^2 + 40 t + 30 = 0. Dividing by 2 we have

-8 t^2 + 20 t + 15 = 0. The quadratic formula gives us

t = [ -20 +- `sqrt( 400 + 480) ] / ( 2 * -8) or

t = 1.25 +- sqrt(880) / 16 or

t = 1.25 +- 29.7 / 16, approx. or

t = 1.25 +- 1.87 or

t = 3.12 or -.62

so now we have substitute these values in

The solution t = 3.12 gives us v(t) = v(3.12) = -32 * 3.12 + 40 = -60, approx.

The interpretation of this result is that when it strikes the ground the balloon is moving downward at 60 feet / second.**

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Self-critique (if necessary):

Ok except I’m not sure which value to use the 59.2 or the -59.2 which come from

-32(-.61)+40=59.52

-32(3.10)+40=-59.2

?????

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Self-critique Rating:

The given solution indicated that the balloon was moving downward at about 60 ft / sec.. I expect you will agree that the direction will be downward.

The upward direction is positive for the given function, because the velocity function uses -32 t for the effect of gravity on the velocity. This implies that the acceleration of gravity is regarded as -32 ft / sec. Since gravity acts downward, the choice of upward as the positive direction is consistent with a negative value for the acceleration of gravity.

Since upward is positive, then, the velocity of the water balloon at the end will be negative.

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Question: How fast is the water balloon moving when it strikes the person's head?

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Your solution:

According to the text the person is 6ft tall so we would like to know the speed of the balloon when it reaches 6 feet. So we substitute into our equation.

-16 t^2 + 40 t + 30=6

-16t^2+40t+24

Once again we use the quadratic equation to solve for t

-b+-squareroot(b^2-4ac)/2a

-40+-squareroot (40^2-4*-16*24)/2*-16

-40+-squareroot(3136)/-32

-40+56/32 -40-56/-32

-.5 3

By taking 3 back to the original equation, V=-32t+40

We get -32*3+40

-56

So the answer is the balloon is falling down at 56ft/sec when it hits the 6ft tall person in the head.

confidence rating #$&*: 3

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Given Solution:

** You have to determine the clock time t when the balloon's altitude is 6 feet. You have the s(t) function. So at what time is the altitude 6 ft?

To answer the question we solve the equation

s(t) = 6, i.e.

-16t^2+40t+30=6. This is a quadratic equation. We rearrange to get

-16t^2+40t+24=0; dividing by -16 we have

t^2-5/2t-3/2=0.

We can solve using the quadratic formula or by factorization, obtaining

t=3 or -.5

When t=3, we have

v(3)=-32*3+40=-56.

Thus the velocity at the 6 ft height is 56 ft/sec downward.

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12:13:49

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Self-critique (if necessary):

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Question: What is the average velocity of the balloon between the two given clock times?

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Your solution:

the 2 given clock times are 1.5 and 3

so you would substitute these two values into the function

V(t)=-32t+40

V(1.5)=-32*1.5+40=-8

V(3)=-32*3+40=-56

To average you add the two together and divide by 2 = -8+-56/2=-32 ft per second is the average velocity.

confidence rating #$&*: 3

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Given Solution:

Average Velocity=-32 m/s

average velocity =

change in position / change in clock time =

(s(3) - s(1.5) ) / (3 sec - 1.5 sec) =

(6 ft - 54 ft) / (1.5 sec) =

-32 ft / sec.

Alternatively since the velocity function is linear, the average velocity is the average of the velocities at the two given clock times:

• vAve = (v(1.5) + v(3) ) / 2 = (-56 ft / sec + (-8 ft / sec) ) / 2 = -32 ft / sec.

This method of averaging only works because the velocity function is linear.

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12:15:31

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Self-critique (if necessary):

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Question: What function describes the position of the balloon as a function of time? How can this function be used to answer the various questions posed in this problem?

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Your solution:

The function that describes the position of the balloon as a function of time is

s(t)=-16t^2+40t+30

this function can be used to answer the question of the vertical velocity of the balloon at different times.

confidence rating #$&*:

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Given Solution:

** On this problem you are given s(0) = 30. So we have

30 = -16 * 0^2 + 40 * 0 + c or

30 = c.

Thus c = 30 and the solution satisfying the initial condition is

s(t) =- 16 t^2 + 40 t + 30.

To find the clock time when the object strikes the ground, note that at the instant of striking the ground s(t) = 0. So we solve

- 16 t^2 + 40 t + 30 = 0,

obtaining solutions t = -.60 sec and t = 3.10 sec.

The latter solution corresponds to our ‘real-world’ solution, in which the object strikes the ground after being released. The first solution, in which t is negative, corresponds to a projectile which ‘peaks’ at height 30 ft at clock time t = 0, and which was at ground level .60 seconds before reaching this peak.

To find when the height is 6 ft, we solve

- 16 t^2 + 40 t + 30 = 6,

obtaining t = -.5 sec and t = 3.0 sec. We accept the t = 3.0 sec solution.

At t = 3.0 sec and t = 3.10 sec the velocities are respectively

v(3.104) = -32 * 3.10 + 40 = -59.2 and

v(3.0) = -32 * 3 + 40 = -56,

indicating velocities of -59.2 ft/s and -56 ft/s at ground level and at the 6 ft height, respectively.

From the fact that it takes .104 sec to travel the last 6 ft we conclude that the average velocity during this interval is

-6 ft / (.104 sec) = -57.7 ft / sec.

This is how we find average velocity. That is, ave velocity is displacement / time interval, vAve = `ds / `dt.

Since velocity is a linear function of clock time, a graph of v vs. t will be linear and the average value of v over an interval will therefore occur at the midpoint clock time, and will be equal to the average of the initial and final velocities over that interval.

In this case the average of the initial and final velocities over the interval during which altitude decreased from 6 ft to 0 is

vAve = (vf + v0) / 2 = (-59.2 + (-56) ) / 2 ft / sec = -57.6 ft / sec.

This agrees with the -57.7 ft / sec average velocity, which was calculated on the basis of the .104 sec interval, which was rounded in the third significant figure. Had the quadratic equation been solved exactly and the exact value ( sqrt(55) + 5 - 3) / 2 of the time interval been used, and the exact corresponding initial and final velocities, the agreement would have been exact.

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12:16:06

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Self-critique (if necessary):

Ok, I understand the different kinds of questions we can answer about the problem as described above.

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Self-critique Rating:

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Question: Problem 3 Section 6.4

6.4.3 (previously 6.4 #12) derivative of (int(ln(t)), t, x, 1).

What is the derivative of this function?

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Your solution:

Use 2nd fundamental calculus theorem to say that the derivative of an antiderivative is just equal to the original function.

So, the original is derivative of (int(ln(t) between x and 1

We thus find derivative of inegeral of ln(x) between c and x where c is a constant

So we get that answer to be ln(x)

confidence rating #$&*:

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Given Solution:

** In the following we'll use the format [int(f(t), t, c, x] to stand for 'the integral of f(t) with respect to t with lower limit c and upper limit x'.

The 2d Fundamental Theorem says that d/dx [ int(f(t)), t, c, x ] = f(x). When applying this Theorem you don't find an antiderivative.

The integral we are given has limits x (lower) and 1 (upper), and is therefore equal to -int(ln(t), t, 1, x). This expression is in the form of the Fundamental Theorem, with c = 1, and its derivative with respect to x is therefore - ln(x).

Note that this Theorem is simply saying that the derivative of an antiderivative is equal to the original function, just like the derivative of an antiderivative of a rate-of-depth-change function is the same rate function we start with. **

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12:23:36

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Self-critique (if necessary):

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Self-critique Rating:3

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Question: Why do we use something besides x for the integrand?

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Your solution:

We use c instead of x because x is the derivative with respect to x and c is the upper limit of the integral. We can’t let x be both of those so we substitute c in for x in the upper integral limit.

confidence rating #$&*:

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Given Solution:

** In the statement of the Theorem the derivative is taken with respect to the variable upper limit of the interval of integration, which is different from the variable of integration.

The upper limit and the variable of integration are two different variables, and hence require two different names. **

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12:24:24

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Self-critique (if necessary):

###Ok, I think my answer above is correct???

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Self-critique Rating:

More correctly, we can't use the same symbol for the upper limit of the integral as we use inside the integrand. Too much possibility of confusion.

Since we're taking the derivative with respect to the upper limit of the integral, that has to be x. So we have to use something else inside the integral.

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Question: Problem 4 Section 6.4.

6.4.4 (previously Section 6.4 #18) derivative of (int(e^-(t^2),t, 0,x^3)

What is the desired derivative?

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Your solution:

We begin with d/dx of integral from 0 to x^3 of e^-t^2

We substitute in a dummy variable for the upper limit of the integral and x for t

d/dx of integral from 0 to z^3 of e^/x^2

now the problem is that the upper limit is not a constant but rather contains a variable

so we now this is a f(g(x) composite function and we’ll have to rename and use chain rule

g=z^3

g’=3z^2

f=e^-x^2

f’ = e^-x^2

for we find g’(x)*f’(x)(g(x))

3z^2*e^z^2^3

So the final answer would be 3x^2*e^-x^2^3

confidence rating #$&*: 2

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Given Solution:

** If we were finding (int(e^-(t^2),t, 0,x) the answer would just be e^-(x^2) by the Second Fundamental Theorem.

However the upper limit on the integral is x^3.

• This makes the expression int(e^(-t^2),t,0,x^3) a composite of f(z) = (int(e^-(t^2),t, 0,z) and g(x) = x^3.

• Be sure you understand how the composite f(g(x)) = f(x^3) would be the original expression int(e^(-t^2),t,0,x^3).

Now we apply the chain rule:

g'(x) = 3x^2, and by the Second Fundamental Theorem f'(z) = e^-(z^2). Thus the derivative is

g'(x) f'(g(x))= 3x^2 * e^-( (x^3)^2 ).

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Self-critique (if necessary):

### My answer differed a little I had 3x^2*e^-x^2^3 where the answer showed 3x^2 * e^-( (x^3)^2 ). Can you tell me what I am doing wrong???

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Self-critique Rating:3

I'll put this into 'chain' notation as df/dz * dz/dx. This should clarify the reasoning.

f(z) = int(e^-(t^2), t from 0 to z)

and

z(x) = x^3

then

dz/dx = 3 x^2,

as you see.

df/dz = e^(-z^2)

df/dx = df/dz * dz/dx = e^(-z^2) * (3 x^2).

Since z = x^3, this becomes

e^(-(x^3)^2) * 3 x^2.

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Question: Find the derivative with respect to x of the integral of e^(t^2) between the limits cos(x) and 3.

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Your solution:

d/dx of intergral of e^t^2 between cos x and 3.

Make substitutions to get d/dx of intergral of e^x^2 between integral of u and 3.

G(x)=cosx

G’=-sinx

F’=e^x^2

Use chain rule

G’(x)*g’(x)(g(x)

-sinx*e^x^2cosx

confidence rating #$&*:

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Given Solution:

If we were finding (int(e^(t^2),t, x, 3) the answer would just be -e^(x^2) by the Second

Fundamental Theorem (along with the reversal of integration limits and therefore sign).

However the lower limit on the integral is cos(x). This makes the expression

int(e^(t^2),t,cos(x), 3) a composite of f(z) = (int(e^(t^2),t, z, 3) and g(x) = cos(x).

Be sure you see that the composite f(g(x)) = f(cos(x)) would be the original expression

int(e^(t^2),t,cos(x),3).

g'(x) = -sin(x), and by the Second Fundamental Theorem f'(z) = -e^(z^2) (again the

negative is because of the reversal of integration limits).

The derivative is therefore

• g'(x) f'(g(x))= -sin(x) * (-e^( (cos(x))^2 ) = sin(x) e^(cos^2(x)).

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Self-critique (if necessary):

###why in the previous problem did we say, “If we were finding (int(e^-(t^2),t, 0,x) the answer would just be e^-(x^2) by the Second Fundamental Theorem.” Yet in this problem we say, “If we were finding (int(e^(t^2),t, x, 3) the answer would just be -e^(x^2) by the Second Fundamental Theorem (along with the reversal of integration limits and therefore sign).” I see the reversal in sign was because of a reversal of intergration limits, how do we know the integration limits will be reversed, is it because the cos of any number is going to be less than 3 so therefore cos would need to be on bottom of the intergral instead of top??? Besides my sign being incorrect because I missed the part about the limit reversal, is my answer ok as written there was a variation in how I wrote the x^2cosx and the way the solution was given???

The statement simply says that the derivative of the integral from x to 3 would be - e^(x^2). It appears that you understand and accept that.

The reason stating that is that the integral in question has cos(x) as the lower limit, and we want to establish that this does reverse the sign of our result.

However x is not one of the limits of the integral, so e^(x^2) would not appear in the final result.

Using 'chain notation' once more:

The function is

f(z) = integral(e^(t^2), t from z to 3)

where

z(x) = cos(x).

df/dz = - e^(z^2)

and

dz/dx = - sin(x).

So df/dx = df/dz * dz/dx = -e^(-z^2) * (-sin(x) = sin(x) e^(z^2) = sin(x) e^(cos^2(x)).

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Self-critique Rating:3

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Question: Section 6.3 Problem 5

6.3.11 was 6.3.5 (was 6.3 #10) dy/dx = 2x+1

What is the general solution to this differential equation?

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Your solution:

Y= X^2+x+c

confidence rating #$&*:

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Given Solution:

student answer: x^2 / 2+x

Instructor response: ** Good. This is an antiderivative of the given function.

So is x^2 + x + c for any constant number c, because the derivative of a constant is zero.

The general solution is therefore the function y(x) = x^2 + x + c . **

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11:18:58

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Self-critique (if necessary):

### I think my solution is correct???

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Self-critique Rating:3

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Question: What is the solution satisfying the given initial condition (part c)?

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Your solution:

Y(0)=12

So Y(0)=0^2+0+c=12

C=12

So answer is x^2 + x + 12

confidence rating #$&*: 3

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Given Solution:

** If for example you are given the initial condition y(0) = 12, then since you know that y(x) = x^2 / 2 + x + c, you have y(0) = 0^2 / 2 + 0 + c = 12.

Thus c = 12 and your particular solution is y(x) = x^2 + x + 12. **

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Self-critique (if necessary):

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Question: What three solutions did you graph, and what does your graph of the

three solutions look like?

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Your solution:

I will choose 3 different values of c

C=-12, c=-6, c=-20

So I have the following to graph

x^2 + x -12

solve by factoring

(x-3)(X+4)

X=3, x=-4

So this graph has points, (3,0) (-4, 0)

x^2 + x -6

solve by factoring

(x-2)(x+3)

X=2, x=-3

So this graph has points (2, 0) (-3, 0)

x^2 + x -20

solve by factoring

(x+5)(x-4)

X=-5, x=4

So this graph has points (-5, 0)(4, 0)

My graphs were 3 parabolas with critical points where they crossed the x axis at x=3, -4, 2, -3, -5, 4

All graphs had inflection points when they crossed the y axis. For negative values of x, the graphs had a decreasing slopes and for positive values of x, the graphs had increasing slope.

confidence rating #$&*: 3

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Given Solution:

** To graph the three solutions you could choose three different values of c.

The graph of x^2 + x is a parabola; you can find its zeros and its vertex using the quadratic formula.

The graph of x^2+ x + c just lies c units higher at every point than the graph of x^2 + x.

So you get a 'stack' of parabolas.

Be sure you work through the details and see the graphs. **

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11:20:51

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Self-critique (if necessary):

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Self-critique Rating:3

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Self-critique (if necessary):

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#*&!

Good work.

You had a little confusion on the second fundamental theorem. I've tried to clarify the order of reasoning by using the 'chain notation' df/dx = df/dz * dz/dx.

Let me know if that doesn't do it.

&&&& I understand much better now, the chain rule was confusing me as was the 2nd fundamental theorem but you made it clear for me by using df/dz*dz/dx. Thank you for your help, I will review this further and practice more.&&&&"

Self-critique (if necessary):

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Self-critique rating:

#*&!

Good work.

You had a little confusion on the second fundamental theorem. I've tried to clarify the order of reasoning by using the 'chain notation' df/dx = df/dz * dz/dx.

Let me know if that doesn't do it.

Self-critique (if necessary):

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Self-critique rating:

#*&!

Good work.

You had a little confusion on the second fundamental theorem. I've tried to clarify the order of reasoning by using the 'chain notation' df/dx = df/dz * dz/dx.

Let me know if that doesn't do it.

Query 2 revised

@& Very good. I was hoping the 'chain' notation would be helpful. Continue to let me know when you have questions. *@

@&
Very good. I was hoping the 'chain' notation would be helpful.

Continue to let me know when you have questions.
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