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1

To find the value for #x#, we must find the point on the graph where #T(x)# is minimal. To do so, we must differentiate #T(x)# and make it equal 0. #"Minimum point": T'(x)=0#

We know from the hint that #T(x)=sqrt(16+x^2)/20+(9-x)/55#

#T'(x)=d/(dx)[(16+x^2)^(1/2)/20+(9-x)/55]#
#color(white)(Xlllll)=d/(dx)[(16+x^2)^(1/2)/20]+d/(dx)[(9-x)/55]#
#color(white)(Xlllll)=(d/(dx)[(16+x^2)^(1/2)])/20+(d/(dx)[9-x])/55#
#color(white)(Xlllll)=(d/(dx)[16+x^2] * 1/2 * (16+x^2)^(1/2-1))/20+(d/(dx)[9]-d/(dx)[x])/55#
#color(white)(Xlllll)=((d/(dx)[16]+d/(dx)[x^2])( 1/2 * (16+x^2)^(-1/2)))/20+(0-1)/55#
#color(white)(Xlllll)=(0+2x * 1/2 * (16+x^2)^(-1/2))/20-1/55#
#color(white)(Xlllll)=(x(16+x^2)^(-1/2))/20-1/55#
#color(white)(Xlllll)=x/(20sqrt(16+x^2))-1/55#

Now we need to make #T'(x)=0#

#x/(20sqrt(16+x^2))-1/55=0#

#x/(20sqrt(16+x^2))=1/55#

#55x=20sqrt(16+x^2)#

#3025x^2=400(16+x^2)#

#3025x^2=6400+400x^2#

#2625x^2=6400#

#x^2=6400/2625=256/105#

#x=sqrt(256/105)=sqrt(256)/sqrt(105)=16/sqrt(105)~~1.56#

2

Answer:

See below.

Explanation:

This differential equation is linear so it can be represented as the sum

#y = y_h+y_p# with

#y'_h+2/x y_h = 0#
#y'_p+2/xy_p = 2 e^(x^2)( (x^2 + 1)/x)#

The solution for #y_h# is easily determined as

#y_h = C_0/x^2#

now supposing that #y_p = (C(x))/x^2# and substituting into

#y'_p+2/xy_p = 2 e^(x^2)( (x^2 + 1)/x)#

we get

#(2 e^(x^2) (1 + x^2))/x - (C'(x))/x^2=0# or

#C'(x) = e^(x^2) x (1 + x^2)# or

#C(x) = e^(x^2)x^2+C_1# and finally

#y = ( e^(x^2)x^2+C_1)/x^2 = C_1/x^2+e^(x^2)#

1

Answer:

See below.

Explanation:

Since the derivative of a function is a function for finding the gradient at a given point, we find this function in a similar way as we would find the gradient of a line. i.e.

#(y_2-y_1)/(x_2-x_1)#

enter image source here

We start with a point #P# on a curve, with coordinates #(x, f(x))#. Then we take a point #Q#, near to #P#, with coordinates #(x+h,f(x+h))# where #h# is a small increment of #x#. We then bring #Q# closer to #P# by allowing #h# to tend towards zero. This then becomes the tangent at #P# and leads to the following:

#dy/dxf(x)=lim_(h->0)(f(x+h)-f(x))/(x+h-x)#

From example we have:

#lim_(h->0)((x+h)^2+1-(x^2+1))/(x+h-x)#

Expanding:

#(x^2+2hx+h^2+1-x^2-1)/(x+h-x)#

Simplifying:

#(2hx+h^2)/(h)#

Cancelling:

#2x+h#

We now take the limit:

#dy/dx(x^2+1)=lim_(h->0)(2x+h)=2x#

1

Answer:

#"see explanation"#

Explanation:

#"let "y=(sinx-cosx)/(sinx+cosx)#

#"differentiate using the "color(blue)"quotient rule"#

#"given "y=(g(x))/(h(x))" then"#

#dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"#

#g(x)=sinx-cosxrArrg'(x)=cosx+sinx#

#h(x)=sinx+cosxrArrh'(x)=cosx-sinx#

#dy/dx=((sinx+cosx)(sinx+cosx)-(sinx-cosx)-1(sinx-cosx))/(sinx+cosx)^2#

#=((sinx+cosx)^2+(sinx-cosx)^2)/(sinx+cosx)^2#

#=(sinx+cosx)^2/(sinx+cosx)^2+(sinx-cosx)^2/(sinx+cosx)^2#

#=1+y^2#

3

Answer:

#n = 115#

Explanation:

Solution 1
Use the formula for an arithmetic series (the sum of an arithmetic sequence).

#S_n = 1/2(n(a_1+a_n))# to get

#LHS = (1/2(n(1+(2n-1))))/(1/2(n(2+2n))) = (2n)/(2+2n) = n/(1+n)#

Now solve #n/(1+n) = 115/116#

Solution 2
The sum of the first #n# odd numbers is #n^2#.

The sum of the first #n# even numbers is twice the sum of the first #n# numbers or #2((n(n+1))/2) = n(n+1)#.

So we have

#LHS = n^2/(n(n+1)) = n/(n+1)# which again leads to solving

#n/(n+1) = 115/116#

1

Answer:

See below.

Explanation:

Exponential decays typically start with a differential equation of the form:

#(dN)/dt prop -N(t)#

That is, the rate at which a population of something decays is directly proportional to the negative of the current population at time #t#. So we can introduce a proportionality constant:

#(dN)/dt=-alphaN(t)#

We will now solve the equation to find a function of #N(t)#:

#->(dN)/N=-alphadt#

#->int(dN)/N=int-alphadt-> ln(N)=-alphat+C#

#->N(t)=Ae^(-alphat)# where #A# is a constant.

This is the general form of the exponential decay formula and will typically have graphs that look like this:

graph{e^-x [-1.465, 3.9, -0.902, 1.782]}

Perhaps an example might help?

Consider a lump of plutonium 239 which initially has #10^24# atoms. After one million years have elapsed years the plutonium now has #2.865times10^11# atoms left. Work out, #A# and #alpha.# When will the plutonium have only #5times10^8# atoms left and what is the decay rate here?

We are told the lump has #10^24# atoms at #t=0# so:

#N(0)=Ae^(0)=10^24-> A=10^24#

Now at 1 million years: #10^6# years:

#N(10^6) = 10^24e^(-alpha(10^6))=2.865times10^11#

Rearrange to get:

#alpha=-1/(10^6)ln((2.865times10^11)/10^24)~~2.888times10^(-5)yr^-1#

So #N(t)=10^(24)e^(-2.888times10^(-5)t)#

For the next part:

#N(t) =5times10^8=10^(24)e^(-2.888times10^(-5)t)#

Rearrange to get #t#:

#t=-1/(2.888times10^(-5))ln((5times10^8)/(10^24))~~1.22times10^6yr#

Now for the last part, the decay rate is already defined a way back at the very start, simply evaluate it at the given time:

#(dN)/dt=-alphat=-2.888times10^(-5)(1.22times10^6)#

#=-35.23# atoms per year.

The idea is to start with differential equation above, which gives the decay rate, and solve it to get the population at any given time.

2

Answer:

#sec(x)tan(x)#

Explanation:

To evaluate #lim_(Deltax to 0)((sec(x+Deltax)-sec(x))/(Deltax))# notice that is of the form #lim_(Deltax to 0)((f(x+Deltax)-f(x))/(Deltax))#, which is the limit definition of the derivative.

So,

#lim_(Deltax to 0)((sec(x+Deltax)-sec(x))/(Deltax))=f'(x)#,

where #f(x)=sec(x)#.

For #f(x)# it's known that #f'(x)=sec(x)tan(x)# so:

#lim_(Deltax to 0)((sec(x+Deltax)-sec(x))/(Deltax))=sec(x)tan(x)#.

1

Answer:

See below.

Explanation:

1)

#ln(cosx)/(4x^2) = ln((cosx)^(1/(4x^2)))#

and #cosx =1-x^2/(2!)+x^4/(4!)-x^6/(6!)+ cdots#

and for small values of #absx# we have

#cosx approx 1-x^2/2# then

#lim_(x->0)ln(cosx)/(4x^2) = ln(lim_(x->0)(1-x^2/2)^(1/(4x^2))) =#

#= ln(lim_(y->0)(1-y)^(1/(8y))) = 1/8#

2)
Making #y=pi-3x rArr x = (pi-y)/3#

#(1-2cosx)/(pi-3x) = (1-(cos(y/3)+sqrt3 sin(y/3)))/y = #

#=(1-cos(y/3))/y-sqrt3 sin(y/3)/y#

now

#lim_(y->0)(1-cos(y/3))/y=0#
#lim_(y->0)-sqrt3 sin(y/3)/y = -sqrt3/3# then

#lim_(x->pi/3)(1-2cosx)/(pi-3x) = -sqrt3/3#

1

Answer:

Please see below.

Explanation:

The standard (basic) definition of convergence of an improper integral tells us that

If #f# is afunction that is not defined at #0#

#int_-a^a f(x) dx# converges if and only if both #int_-a^0 f(x) dx# and #int_0^a f(x) dx# converge.

There is another definition, called the Cauchy Principle Value. For the integral you are working on,
the Cauchy Principle Value is

#lim_(epsilonrarr0^+)[int_-1^(0-epsilon)1/x^3 dx + int_(0+epsilon)^1 1/x^3 dx] = 0#

The standard (or Basic) definition requires us to take the two limit separately. The Cauchy principal value takes the limit of the sum.

As for you specific question, I am not confident about any definition of convergence. that would allow us to answer the question, "Does #lim_(ararr0^+)(1/a^2)+lim_(brarr0-)(-1/b^2)# converge?"

I can say that in the extended real numbers this evaluates to #oo+(-oo)# which is not defined in the extended reals. So I would call that divergent.

1

Answer:

Yes, we do multiply by the conjugate, but it is easy to be mistaken about what the conjugate is.

Explanation:

The conjugate of #x^(1/2)-y^(1/2)# is #x^(1/2)+y^(1/2)# (and vice versa) because

#(a-b)(a+b) = a^2-b^2#.

So if either of #a# or #b# involves a square root, then its square does not involve a square root..

The analog for cube roots is based on the fact that

#(a-b)(a^2+ab+b^2) = a^3-b^3#.

So the conjugate of #x^(1/3)-y^(1/3)# #" "# is #" "# #x^(2/3)+(xy)^(1/3) + y^(2/3)# (and vice versa)

#(((x+h)^(1/3) - x^(1/3))) / h * (((x+h)^(2/3)+(x(x+h))^(1/3) + x^(2/3)))/ (((x+h)^(2/3)+(x(x+h))^(1/3) + x^(2/3))) = ((x+h)-x)/( h((x+h)^(2/3)+(x(x+h))^(1/3) + x^(2/3))#

# = 1/((x+h)^(2/3)+(x(x+h))^(1/3) + x^(2/3))#

And, now we can evaluate the limit

# lim_(hrarr0) 1/((x+h)^(2/3)+(x(x+h))^(1/3) + x^(2/3)) = 1/(3x^(2/3))#