Help with related rates?

10)     Okay Chris I have accepted your challenge.

I had to really think!

Now we know that    $$\frac{dx}{dt}=5$$

$$\\sin\theta=\frac{x+y}{15}=\frac{y}{6}\\\\
\begin{array}{rll}
\frac{x+y}{15}&=&\frac{y}{6}\\\\
6(x+y)&=&15y\\
6x+6y&=&15y\\
6x&=&9y\\
y&=&\frac{6x}{9}=\frac{2x}{3}\\
\frac{dy}{dx}&=&\frac{2}{3}\\
\end{array}$$

Now we need to find      $$\frac{dy}{dt}$$

$$\\\frac{dy}{dt}=\frac{dy}{dx}\times \frac{dx}{dt}\\\\
\frac{dy}{dt}=\frac{2}{3}\times 5\\\\
\frac{dy}{dt}=\frac{10}{3}\\\\$$

So the tip of the shadow is moving away at the constant speed of  10/3   feet per second

I hope that is correct.  (I think that it is)     

Here's (1)

y = x^2 + x - 2

y' = 2x + 1

Set this to 0

2x + 1 = 0

2x = -1

x = -1/2

And this is the critical point.

Using the second derivative, we have

y'' = 2

This indicates that the curve is concave up at every point.  Thus, the curve is at an absolute minimum at x = -1/2. And since this polynomial is continuous everywhere, the curve is decreasing on (-∞, -1/2) and  increasing on (-1/2, ∞)

Here's (2)

y = 5x^2 + 2x

y' = 10x + 2

Set this to 0

10x + 2 = 0

10x = -2

x = -(1/5)

Note that the second derivative is just

y'' = 2

This curve is continuous and concave up at all points so we have a minimum at  x = -(1/5)

Here's (3)

y = x^5 - 20x^2

y' = 5x^4 - 40x

Set this to 0

5x^4 - 40x =  0      factor 

5x(x^3 - 8) = 0    divide by 5 on both sides

x(x^3 - 8) = 0

So x = 0 is one critical point

x^3 = 8      take the cube root of both sides

x = 2     and this is another critical point

And taking the second derivative we have

y'' = 20x^3 - 40

And putting 0 into this we have a negative.....so we have a max when x = 0

And putting 2 into this we have a positive......so we have a min at x = 2

Here's (4)

y = x^4 - 4x^3

y' = 4x^3 - 12x^2

y'' = 12x^2 - 24x

Set this to 0

12x^2 - 24x = 0     factor

12x ( x - 2) = 0

And x = 0  and x = 2 are possible inflection points.

Test a point on either side of 0 in the second derivative (I'll use -1 and 1)......putting -1 into the second derivative we have a positive and putting 1 into the second derivative gives a negative.....so x = 0 is an inflection point.

And putiing 3 into the second derivative gives a positive (and since 1 gives a negative), then x = 2 is another inflection point.

Here's the graph of this one.......https://www.desmos.com/calculator/wgm5x0ifno

Here's (5)

y = 2x^3 - 3x^2 - 12x

y' = 6x^2 - 6x - 12

y'' = 12x - 6

Set this to 0

12x - 6 = 0    factor

6 (2x -1) = 0       divide by 6 on both sides

2x - 1 = 0

x = 1/2        and this is a possible inflection point

Test a point on either side of x = 1/2 in the second derivative (I'll pick 0 and 1)

2(0) - 1 = -1

2(1) -1 = 1

So, since the second derivative has different values on either side of x = 1/2......this is an inflection point.

Here's the graph........https://www.desmos.com/calculator/ogg8dusr6b

Here's (6)

y = 2 + 3x - x^3

y ' = 3 - 3x^2

y'' = -6x

Notice that this will be concave down on ( 0, ∞) because at all x's in this interval, the second derivative will be negative

Here's the graph.........https://www.desmos.com/calculator/2bbqxjmorn

Here's (7)

Let x be the width of the garden

The, the amount of fencing available for the lengths is 300-2x.....and since there are 3 equal "lengths," the length of one is just (300 - 2x)/3

So....the area is

A = (x)((300 - 2x) / 3)) = 100x - (2/3)x^2

A' = 100 - (4/3)x

Set this to 0

100 - (4/3)x = 0

(4/3)x = 100

x = 75

Taking the second derivative, we have

y = -(4/3)....so this curve is concave down everywhere....so x = 75 is a max

And the area is (75)(300-2(75))/3  =  (75)(150/3) = (75)(50) = 3750 ft²

Here's the graph.......https://www.desmos.com/calculator/k77tcaunrm

Here's 8

Let x be the width parallel to the river...so the amount remaing for two equal sides is  b - x ....... so...the length of one side is just(b - x) /2

So the area is

A = x ( (b - x) / 2 )   = (1/2)bx - (1/2)x^2

A ' = (1/2)b - x      set to 0   and we have

x = (1/2)b

And length of the side is  (b - (1/2)b) / 2  = (1/2)b / 2  = (1 / 4) b

And the max area is just  (1/2)b * (1 /4)b =  (1/8)b^2

Here's (9)

We can use the idea of a right triangle for this one.  We have

r^2 = x^2 + y^2       and differentiating this with respect to time (t), we have...

2r(dr/dt) = 2x(dx/dt) + 2y(dy/dt)

We are looking for dr/dt

dx/dt  = 500mi/h      dy/dt = 0   (the plane is not rising or falling)......so we can ignore the last term above..... and r = 2  and x = 1 at the time of interest .......so we have

2(2mi)(dr/dt) = 2(1mile)(500 m/hr)

4mi(dr/dt) = 1000m²/hr        divide both sides by (4mi)

(dr/dt) = (1000m²/hr) / (4mi)   =  250 m/hr

WOW Chris - you have been a busy wolf !     

@Melody....Yes, I have......I might need you to do (10)....I never can quite remember what to do here.....I remember that it's a similar triangle type problem, and I know where to find a similar problem, but I think you can do it off the top of your head......!!!!

I believe I can handle (11)........

Omg.. you guys are amazing.. seriously.

Here's (11)

Let the side of the square cut out = x

So, if you can imagine it....the area of the bottom of the box will be (20 -2x)(20-2x)....and the height of the box = x

So....the volume will be

V = x*(20 - 2x)^2    ..... let's just expand this out

V = x(4 x^2 - 80x + 400)

V = 4x^3 - 80x^2 + 400x

V' = 12x^2 - 160x + 400    divide through by 4 

V' = 3x^2 - 40x + 100   let's see if this will factor (and set it to 0)

(3x -10)(x - 10) = 0

Reject the second factor...it would make a side length at the bottom = 0 !!

So we have

3x - 10 = 0

3x = 10

x = 10/3  ....  this is the height

And the length of each side of the bottom = (20 - 2(10/3)) = (20 - 20/3) = (60/3 - 20/3) =40/3

And the volume is just...h x area of the base = (10/3)(40/3)^2  = about 592.6 cm^3

Here's a graph......https://www.desmos.com/calculator/zq8xlob32o

That looks nice, Melody.

Thanks so much :D

You guys rock!

They rock indeed!!!!