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1

Answer:

BTW, the graph of this equation is very beautiful

Explanation:

Here's the graph:

enter image source here

1

Answer:

The answer is #=1/5#

Explanation:

This is an improper integral

Start by calculating the definite integral

Let #u=x^5#, #=>#, #du=5x^4dx#

Therefore,

#intx^9e^(-x^5)dx=1/5intue^(-u)du#

Perform the integration by parts

#p=u#, #=>#, #p'=1#

#q'=e^-u#, #=>#, #q=-e^-u#

Therefore,

#intpq'=pq-intp'q#

#1/5intue^(-u)du=1/5(-ue^-u+inte^-udu)=1/5(-ue^-u-e^-u)#

#=-e^-(x^5)/5(x^5+1)#

Therefore,

#lim_(t->+oo)int_0^tx^9e^(-x^5)dx=lim_(t->+oo) [[ -e^(-t^5))/5(t^5+1)]_0^oo#

#=lim_(t->+oo)(-e^(-t^5)/5(t^5+1))+1/5#

#lim_(t->+oo)((t^5+1)/(5e^(t^5)))+1/5#

#=0+1/5#

The integral converges

1

Answer:

#0#

Explanation:

Making #x = n^2#

#lim_(x->oo) ((x + 3)/(2 x + 4))^x =lim_(x->oo)( (1+3/x)/(2+4/x))^x# and for #x# very large

# (1+3/x)/(2+4/x) approx 1/2# then

#lim_(x->oo) ((x + 3)/(2 x + 4))^x = lim_(x->oo)(1/2)^x = 0#

1

Answer:

#lim_(x->0) (tanx-sinx)/x^3 = 1/2#

Explanation:

Transform the function in this way:

#(tanx-sinx)/x^3 = 1/x^3(sinx/cosx-sinx)#

#(tanx-sinx)/x^3 = 1/x^3((sinx-sinxcosx)/cosx)#

#(tanx-sinx)/x^3 = sinx/x^3(1-cosx)/cosx#

#(tanx-sinx)/x^3 = (sinx/x)((1-cosx)/x^2)( 1/cosx)#

We can use now the well known trigonometric limit:

#lim_(x->0) sinx / x = 1#

and using the trigonometric identity:

#sin^2alpha = (1-cos2alpha)/2#

we have:

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

While the third function is continuous so:

#lim_(x->0) 1/cosx = 1/1 = 1#

and we can conclude that:

#lim_(x->0) (tanx-sinx)/x^3 = lim_(x->0) (sinx/x)((1-cosx)/x^2)( 1/cosx) = 1 xx 1/2 xx 1 = 1/2#

graph{(tanx-sinx)/x^3 [-1.25, 1.25, -0.025, 1]}

1

Answer:

See the proof below

Explanation:

Let #vecu= < a, b,c ># in the basis #beta=(hati, hatj, hatk)#

and #vecv= < p,q,r > #

It is given that #||vecu|| = ||vecv||#

Then

#vecu+vecv=< a, b,c > + < p,q,r > #

#=< a+p,b+q,c+r >#

#vecu-vecv=< a, b,c > - < p,q,r > #

#=< a-p, b-q, c-r >#

The dot product is

#(vecu+vecv).(vecu-vecv)#

#=< a+p,b+q,c+r > . < a-p, b-q, c-r >#

#= (a+p)(a-p) +(b+q)(b-q) + (c+r)(c-r) #

#=a^2-p^2+b^2-q^2+c^2-r^2#

#=(a^2+b^2+c^2) - (p^2+q^2+r^2)#

#=||vecu||^2- ||vecv||^2=0#

Therefore,

#(vecu+vecv)# and #(vecu-vecv)# are orthogonal since their dots products #=0#

1

Answer:

The function is concave down on #(-oo, oo)#.

Explanation:

Concavity is determined by the second derivative.

#f''(x) = -8#

This is negative on all #x#, thus it will be concave down on all #x#.

Hopefully this helps!

2

Answer:

#lim_(x→∞) f(x) = 5#

Explanation:

#lim_(x→∞) (5x^2-3x+4)/(x-1)^2 = #

#lim_(x→∞) (5x^2-3x+4)/(x^2-2x+1) = #

#lim_(x→∞) (5x^2((5x^2)/(5x^2)-(3x)/(5x^2)+4/(5x^2)))/(x^2(x^2/x^2-(2x)/x^2+1/x^2)) = #

#lim_(x→∞) (5x^2(1-(3)/(5x)+4/(5x^2)))/(x^2(1-(2)/x+1/x^2)) = #

#lim_(x→∞) (5x^2(1-0+0))/(x^2(1-0+0)) = (5x^2)/x^2 = 5#

1

Answer:

#(d theta)/(dt) = 37/20#

Explanation:

You were on the correct track with your setup on number 8, but when you took the derivative of the #y/z# term on the right side with respect to time #t#, you did not use the Quotient Rule to perform the derivative, which rendered the rest of the problem incorrect.

The proper setup is #sin theta = y/z# as you noted. From there:

#(cos theta) (d theta)/(dt) = (z*(dy)/(dt) - y*(dz)/(dt))/(z^2)#

It is difficult to read the image included, but I believe the setup involves #y = 8#, #z = 10#, #(dy)/(dt) = 10#, #(dz)/(dt) = -6#, and #cos theta = 8/10#, which means you now have:

#(cos theta) (d theta)/(dt) = (z*(dy)/(dt) - y*(dz)/(dt))/(z^2)#

#(8/10) (d theta)/(dt) = (10(10) - 8(-6))/(10^2)#

#(4/5) (d theta)/(dt) = (100+48)/100#

#(4/5) (d theta)/(dt) = 148/100#

#(d theta)/(dt) = 37/25 * 5/4 = 37/20#

A quick sanity check on the sign of the result can help catch errors as they are in progress. As the ships continue moving, ship A will get closer to the lighthouse, while ship B will move farther away.

This action will serve to make the angle #theta# become larger and larger, much like how the angle made by a ladder to the ground increases as you push a ladder up the side of a house. (By the way, this is a common alternative word problem that tests the exact same math in many Calculus classes.)

Your original attempt ended with a negative answer, which would indicate the angle is decreasing - a result that contradicts the nature of the situation. It might not help you find the issue, but it can at least give you a hint that you're on the wrong track somewhere.

1

Answer:

#(dP)/(dt) = 2/5nR + 4#

Explanation:

In the Ideal Gas Law formula, there are 3 variables and 2 constants. #P# represents the pressure (in this problem in units of pascals), #V# represents the volume (in this problem in units of #m^3#), and #T# represents the temperature (in this problem in Kelvin). We consider #n# to be the amount of gas present (usually in moles), while #R# is the Ideal Gas Constant, or a fixed value.

This is a "classic" related rates problem. We need to take derivatives of every variable (#P, V, T#) with respect to time #t#. To do so, we perform "traditional" derivatives on each of those terms in the equation, but we ensure that we include a #(dP)/(dt)#, #(dV)/(dt)#, or #(dT)/(dt)# term as necessary.

#PV = nRT#

#Punderbrace((1*(dV)/(dt)))_ "d(V)" + Vunderbrace((1*(dP)/(dt)))_ "d(P)" = nRunderbrace((1*(dT)/(dt)))_ "d(T)"#

#P((dV)/(dt)) + V((dP)/(dt)) = nR((dT)/(dt))#

We have all of the values we need provided to us to substitute into this related rates expression:

#{(P,20 "pascals","Pressure"),((dV)/(dt),-2 m^3/min,"Volume Change"),(V,10 m^3,"Volume"),((dP)/(dt),?,"Pres. Change"),((dT)/(dt),4 K/min,"Temp Change"):}#

#(20)(-2) + (10)((dP)/(dt)) = nR(4)#

#-40 + 10((dP)/(dt)) = 4nR #

#10((dP)/(dt)) = 4nR + 40 #

#(dP)/(dt) = (4nR + 40)/10 = 2/5nR + 4#

A quick "sanity check" verifies that the sign of our answer is what we'd expect. Since P is inversely related to V and directly related to T, if the volume is decreasing, and the temperature is increasing, we'd definitely expect the pressure P to be increasing (positive). Since #n# and #R# are both positive constants, the final expression will be positive.

1

Answer:

#x^2/2*Ln[(x+1)/(1-x)]-x^2/2-1/2*Ln(x^2-1)+C#

Explanation:

#int x*ln[(x+1)/(1-x)]*dx#

After setting #dv=x*dx# an #u=Ln[(x+1)/(1-x)]=Ln(x+1)-Ln(1-x)#,

#v=x^2/2# and #du=(dx)/(x+1)-(-dx)/(1-x)=(dx)/(x+1)+(dx)/(x-1)=(2x*dx)/(x^2-1)#

So,

#int u*dv=uv-int v*du#

#int x*ln[(x+1)/(1-x)]*dx=x^2/2*Ln[(x+1)/(1-x)]-int x^2/2*(2x*dx)/(x^2-1)#

=#x^2/2*Ln[(x+1)/(1-x)]-int (x^3*dx)/(x^2-1)#

=#x^2/2*Ln[(x+1)/(1-x)]-int ((x^3-x+x)*dx)/(x^2-1)#

=#x^2/2*Ln[(x+1)/(1-x)]-int ((x^3-x)*dx)/(x^2-1)-int (x*dx)/(x^2-1)#

=#x^2/2*Ln[(x+1)/(1-x)]-int x*dx-1/2*int (2x*dx)/(x^2-1)#

=#x^2/2*Ln[(x+1)/(1-x)]-x^2/2-1/2*Ln(x^2-1)+C#