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Find properties of rates of change of x^3 + y^3 = 1

by RoRi

Given the equation

    \[ x^3 + y^3 = 1. \]

This defines y as a function of x.

  1. Without solving for y show that the derivative y' satisfies the equation:

        \[ x^2 + y^2 y' = 0. \]

    (Assume the derivative y' exists.)

  2. Show that

        \[ y'' = -2xy^{-5} \]

    whenever y \neq 0. (Assume the second derivative y'' exists.)

  1. For this part we differentiate each side with respect to x (keeping in mind that y is a function of x, so we need to use the chain rule to differentiate y^3).

        \begin{align*} x^3 + y^3 = 1 && \implies && 3x^2 + 3y^2 y'&= 0 \\ && \implies && x^2 + y^2 y' &= 0. \end{align*}

  2. Using part (a) we differentiate y' to find y'',

        \begin{align*} &&y' &= -\frac{x^2}{y^2}\\ \implies && y'' &= \frac{-2xy^2 + x^2 2y y'}{y^4} \\ &&&= \frac{-2xy^2 + x^2 2 y \left( -\frac{x^2}{y^2} \right)}{y^4} \\ &&&= \frac{-2xy^3 - 2x^4}{y^5} \\ &&&= \frac{-2x(y^3 + x^3)}{y^5} \\ &&&= -2xy^{-5} &(\text{since } x^3 + y^3 = 1). \end{align*}

Find the rate of change in volume of a right circular cylinder

by RoRi

Given a right circular cylinder whose radius increases at a constant rate and whose altitude is a linear function of the radius. Also, given that the altitude is increasing at a rate three times that of the radius. The volume is increasing at a rate of 1 cubic feet per second when the radius is 6 feet. When the radius is 36 feet the volume is increasing at a rate of n cubic feet per second. Find the value of the integer n.

The following diagram illustrates the setup:

Rendered by QuickLaTeX.com

Since the altitude (which we denote h) is a linear function of the radius and increases three times as quickly, we have

    \[ h = 3r + c \implies \frac{dh}{dr} = 3. \]

When r=1, we are given h=6. Thus, we solve for c and get c = 3.
When r=6, we have

    \[ \frac{dV}{dt} = 1 \text{ cubic foot per second}. \]

So, since r = 6 implies h = 3(6) + 3 = 21 and

    \[ V = \pi r^2 h \implies \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} + 2 \pi r h \frac{dr}{dt} = 1, \]

we have,

    \[ \pi (36) \left( 3 \frac{dr}{dt} \right) + 2 \pi (6) (21) \left( \frac{dr}{dt} \right) = 1. \]

Solving for \frac{dr}{dt} we obtain,

    \[ \frac{dr}{dt} = \frac{1}{360 \pi}. \]

Then, when r = 36 we have

    \begin{align*} n &= \pi (36)^2 \left( \frac{1}{120 \pi} \right) + 2 \pi (36) (111) \left( \frac{1}{360 \pi} \right) &(r = 36 \implies h = 111) \\ &= \frac{54}{5} + \frac{111}{5} \\ &= 33. \end{align*}

Find the rate of change of the area of a triangle as a vertex moves

by RoRi

Let ABC be a right triangle in the plane. The angle at vertex B is the right angle, the vertex A is fixed at the origin, and the vertex C lies on the parabola y = 1 + \frac{7}{36} x^2. At time t = 0, the vertex B is at the point (0,1) and moves up the y-axis at a constant rate of 2 centimeters per second. What is the rate of change of the area of the triangle at time t = \frac{7}{2} seconds?

Here’s a graph of the setup:

Rendered by QuickLaTeX.com

We are given that the vertex B moves upward along the y-axis at a constant rate of 2 centimeters per second, this means,

    \[ \frac{dy}{dt} = 2. \]

Then, we compute the area of the triangle, call it A_{ABC}, in terms of x and y,

    \[ A_{ABC} = \frac{1}{2} xy \quad \implies \quad \frac{d A_{ABC}}{dt} = \frac{y}{2} \frac{dx}{dt} + \frac{x}{2} \frac{dy}{dt}. \]

(Where we used the product rule to differentiate with respect to t, recalling that both x and y are functions of t as well.) We are then given the formula for y with respect to x,

    \begin{align*} &&y &= 1 + \frac{7}{36} x^2 \\ \implies && \frac{dy}{dt} &= \frac{14}{36} x \frac{dx}{dt} \\ \implies &&& = \frac{7x}{18} \frac{dx}{dt} \\ \implies && \frac{dx}{dt} & = \frac{18}{7x} \frac{dy}{dt} \\ \implies &&& = \frac{36}{7x}. \end{align*}

So, when t = \frac{7}{2} we have y = 8 (see the comments for an explanation of how we calculated y) which implies x = 6. Therefore,

    \[ \frac{d A_{ABC}}{dt} = 4 \frac{dx}{dt} + 3 \frac{dy}{dt} = \frac{24}{7} + 6 = \frac{66}{7}. \]

Find the relative rate of change of volume to rate of change of height in a water tank

by RoRi

Consider a water tank shaped like a hemisphere with a radius of 10 feet. At time t let,

    \begin{align*} h &= \text{ depth of the water in the tank}, \\ r &= \text{ radius of the surface of the water}, \\ V &= \text{ volume of water in the tank}. \end{align*}

Find the rate of change of the volume relative to the rate of change of the height (\frac{dV}{dh}) when h = 5.
If water is flowing into the tank at a constant rate of 5 \sqrt{3} cubic feet per second, find \frac{dr}{dt} when h = 5.

For this problem we will consider the graph of the following function (the hemisphere tank and water will then be obtained as solids of revolution of this graph about the y-axis):

Rendered by QuickLaTeX.com

First, we find a formula for the volume of the water in the tank as a function of h. We do this by considering the solid of revolution (for a review of solids of revolution see these exercises) of \sqrt{20y-y^2} about the y-axis:

    \begin{align*} V & = \pi \int_0^h \left( \sqrt{20y - y^2} \right)^2 \, dy \\ &= \pi \int_0^h (20y -y^2) \, dy \\ &= \pi \left( 10y^2 - \frac{y^3}{3} \right)\Bigr \rvert_0^h \\ &= 10 \pi h^2 - \frac{\pi h^3}{3}. \end{align*}

Differentiating with respect to h we then have,

    \[ \frac{dV}{dh} = 20 \pi h - \pi h^2. \]

So, when h = 5 feet we have

    \[ \frac{dV}{dh} = 75 \pi. \]

Next, we are given \frac{dV}{dt} = 5 \sqrt{3} cubic feet per second. We know from above, \frac{dV}{dh} = 20 \pi h - \pi h^2. So,

    \[ \frac{dV}{dh} \cdot \frac{dh}{dt} = \frac{dV}{dt} \quad \implies \quad \frac{dh}{dt}= \frac{5 \sqrt{3}}{20 \pi h - \pi h^2}. \]

Then, to get r in terms of h we evaluate,

    \begin{align*} r^2 + (10-h)^2 = 100 \quad &\implies \quad r = \sqrt{20h - h^2} \\ &\implies \frac{dr}{dh} = \frac{10-h}{\sqrt{20h - h^2}}. \end{align*}

Thus,

    \[ \frac{dr}{dt} = \left( \frac{dh}{dt} \right) \left( \frac{dr}{dh} \right) = \left( \frac{5 \sqrt{3}}{20 \pi h - \pi h^2} \right) \left( \frac{10-h}{\sqrt{20h - h^2}} \right). \]

So, if h = 5, we have

    \[ \frac{dr}{dt} = \left( \frac{\sqrt{3}}{15 \pi} \right) \left( \frac{5}{\sqrt{75}} \right) = \frac{1}{15 \pi} \text{ feet per second}. \]

Use related rates to find the rate at which we need to add water to a leaky tank

by RoRi

Given a water tank shaped like a right circular cone with radius 15 feet at its base and altitude 10 feet. Water leaks from the tank at a constant rate of 1 cubic foot per second, and water is added to the tank at a constant rate of c cubic feet per second. Find the value of c so that the water level will be rising at a rate of 4 feet per second when the depth of the water is 2 feet.

The following diagram illustrates the situation:

Rendered by QuickLaTeX.com

First, we compute the radius of the water level in terms of the height of the water,

    \[ r = \frac{15}{10} h = \frac{3}{2} h \quad \implies \quad \frac{dr}{dt} = \frac{3}{2} \frac{dh}{dt}. \]

Furthermore, the radius at the water line is 3 feet when the height of the water is 2 feet. So, computing the volume of water in terms of r and h we have,

    \begin{align*} V &= \frac{1}{3} \pi r^2 h \\ \implies \frac{dV}{dt} &= \frac{2}{3} \pi r h \frac{dr}{dt} + \frac{1}{3} \pi r^2 \frac{dh}{dt}. \end{align*}

(As in the previous exercise, we are careful to use the product rule when evaluating this derivative.)
The problem then gives us the rate of change of volume of the water is

    \[ \frac{dV}{dt} = (c-1) \text{ cubic feet per second}. \]

(Since we are adding c cubic feet per second, and 1 cubic foot per second is leaking out.) So we set \frac{dh}{dt} = 4 and solve for c,

    \begin{align*} (c-1) &= \frac{2}{3} \pi (3)(2) \left( \frac{3}{2} \right) (4) + \frac{1}{3} \pi (9)(4) = 36 \pi \\ \implies c&= 36 \pi + 1 \text{ cubic feet per second}. \end{align*}

Use related rates to find the water level in a tank shaped like a right circular cone

by RoRi

Given a water tank in the shape of a right circular cone with radius of the base 4 feet and altitude of 10 feet. Water is added to the tank at a constant rate of 5 cubic feet per minute. Find the rate at which the water level is rising when the depth of the water is 5 feet when:

  1. the vertex of the cone is pointed up;
  2. the vertex of the cone is pointed down.
  1. First, we give the general setup for the problem. The following diagram may be helpful,

    Rendered by QuickLaTeX.com

    Next, we compute the radius of the waterline in terms of the height (or altitude) h,

        \[ r = -\frac{4}{10}h = -\frac{2}{5} h \quad \implies \quad \frac{dr}{dt} = -\frac{2}{5} \frac{dh}{dt}. \]

    Then, using the formula for the volume of a right circular cone we have and letting V be the volume of the water we have V is equal to the volume of the entire tank less the volume of the empty (smaller) cone above the water:

        \begin{align*} V &= \frac{1}{3} \pi (4^2) (10^2) - \frac{1}{3} \pi r^2 (10-h) \\ &= \frac{1600 \pi}{3} - \frac{10}{3} \pi r^2 + \frac{1}{3} \pi r^2 h \\ &= \frac{1600 \pi}{3} - r^2 \left( \frac{10 \pi}{3} - \frac{\pi}{3} h \right). \end{align*}

    This implies

        \[ \frac{dV}{dt} = -2r \left( \frac{10 \pi}{3} - \frac{\pi}{3} h \right) \frac{dr}{dt} - r^2 \left( - \frac{\pi}{3} \frac{dh}{dt} \right). \]

    (Here, we needed to use the product rule for derivatives since both r and h are functions of t, so when we differentiate their product we have be careful to use the product rule and get both terms shown above.)

    The problem gives us that the rate of change in the volume of water is 5 cubic feet per minute, so,

        \[ \frac{dV}{dt} = 5. \]

    When h = 5, we have r = 2. Substituting these values and \frac{dr}{dt} = -\frac{2}{5} \frac{dh}{dt} obtained above we have,

        \begin{align*} 5 &= -2 (2) \left( \frac{10 \pi}{3} - \frac{\pi}{3} (5) \right) \left( -\frac{2}{5} \frac{dh}{dt} \right) - (2^2) \left( -\frac{\pi}{3} \frac{dh}{dt} \right) \\[9pt] &= \frac{-20 \pi}{3} \frac{-2}{5} \frac{dh}{dt} + \frac{4 \pi}{3} \frac{dh}{dt} \\[9pt] &= \frac{60 \pi}{15} \frac{dh}{dt} \\[9pt] \implies \frac{dh}{dt} &= \frac{5}{4 \pi}. \end{align*}

  2. Part (b) will be almost the same as part (a) with a few minor changes. Here is the diagram we will work from,

    Rendered by QuickLaTeX.com

    Now, the radius of the waterline in terms of the height of the water h is

        \[ r = \frac{4}{10} h = \frac{2}{5} h \quad \implies \quad \frac{dr}{dt} = \frac{2}{5} \frac{dh}{dt}. \]

    Then, again using the volume of a right circular cone and letting V denote the volume of the water (things are a bit simpler this time since the water is a cone, and we don’t have to subtract anything) we have

        \[ V = \frac{1}{3} \pi r^2 h \quad \implies \quad \frac{dV}{dt} = \frac{2}{3} \pi r h \frac{dr}{dt} + \frac{1}{3} \pi r^2 \frac{dh}{dt}. \]

    (As in part (a), we had to be careful to use the product rule since both h and r are functions of t). The problem then gives us that the rate of change in the volume of water is 5 cubic feet per minute, so again,

        \[ \frac{dV}{dt} = 5. \]

    When h = 5 we still have r = 2. Substituting these values along with \frac{dr}{dt} = \frac{2}{5} \frac{dh}{dt} we have

        \begin{align*} 5 &= \frac{2}{3} \pi (2) (5) \left( \frac{2}{5} \frac{dh}{dt} \right) + \frac{1}{3} \pi (2^2) \frac{dh}{dt} \\[9pt] &= \frac{8 \pi}{3} \frac{dh}{dt} + \frac{4 \pi}{3} \frac{dh}{dt} \\[9pt] &= \frac{12 \pi}{3} \frac{dh}{dt} \\[9pt] \implies \frac{dh}{dt} &= \frac{5}{4 \pi}. \end{align*}

    The same answer as in part (a).

How fast is a boat approaching a lighthouse

by RoRi

Given a boat sailing at a constant speed of 12 miles per hour. The boat is 4 miles offshore, sailing parallel to a straight coast. How fast is it approaching a lighthouse on the shore when it is exactly 5 miles from the lighthouse?

Let

    \begin{align*} y &= \text{ the distance along the shore to the lighthouse}. \\ x &= \text{ the distance from the boat to the lighthouse}. \end{align*}

The following diagram illustrates the setup:

Rendered by QuickLaTeX.com

The problem gives us that the distance along the shore to the lighthouse is changing at a rate of 12 miles per hour (since the boat is moving at 12 miles per hour and stays parallel to the straight shoreline). Thus, \frac{dy}{dt} = 12. Then, using the Pythagorean theorem, when x = 5 we have y = 3. Furthermore, solving x in terms of y and differentiating we have

    \[ x = \sqrt{y^2 + 16} \quad \implies \quad \frac{dy}{dx} = \frac{y}{\sqrt{y^2 + 16}}. \]

So, we then have at y = 3,

    \[ \frac{dx}{dt} = \frac{dy}{dt} \cdot \frac{dx}{dy} = (12) \left( \frac{3}{\sqrt{9+16}} \right) = \frac{36}{5} = 7.2 \text{ mph}. \]