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

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:

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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,

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

3 comments

  1. Daniel says:

    Hi RoRi,

    I’m self-teaching calculus directly from Apostol’s book(s) and it has been a great and rewarding slog. But although I get approx 95% correct, he sets some problems that just floor me. Like this one here.

    Excuse me if what I’m asking is doltishly simple – it is always possible I’ve missed out on something pretty basic – but I can’t get my head around why:

    If Area = 1/2xy then dA/dt = 1/2y dx/dt + 1/2x dy/dt.

    Why not 1/2(dx/dt dy/dt)? Where does the addition function come from (and I know it leads to the right answer)?

    I tried to work it out from basic principles using the simplified example of expanding rectangular areas (where dx and dy are easily represented and drawn) and just can’t see where the “+” fits in.

    Honestly, apologies if this is dismally basic but I’ve been stuck on the question for weeks and really want to move on. Oh and thanks for your resource. I will try to use it as little as possible!

    Daniel

    • RoRi says:

      Yeah, we start at the point (0,1) so the initial value of y is y_0 = 1. Then we are told that the point moves up the y-axis at a constant rate of 2 centimeters per second, so the ending value of y is

          \begin{align*}  y &= y_0 + 2\cdot t \\  &= 1 + 2 \cdot \frac{7}{2} \\  &= 8. \end{align*}

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