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The sides x and y of a right-angled triangle are 2 mm and 3 mm respectively with the hypotenuse as 'r'. Calculate the approximate change in length of the hypotenuse if x changes to 3 mm and y changes to 5 mm. Given that x = t − t2 and y = ln(3t), determine: dy/dx, d2y/dx2

Mathematics

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Answer

81313 mm≈2.22 mm\frac{8\sqrt{13}}{13} \text{ mm} \approx 2.22 \text{ mm}

dydx=1t(1−2t)\frac{dy}{dx} = \frac{1}{t(1 - 2t)}

d2ydx2=−1−4tt2(1−2t)3\frac{d^2 y}{dx^2} = -\frac{1 - 4t}{t^2(1 - 2t)^3}

Graph of y = ln(3t)

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Key points of y = ln(3t)
Pointty
x-intercept0.330

Step-by-step solution

9 steps

1.1 The sides xx and yy of a right-angled triangle are 2 mm and 3 mm respectively with the hypotenuse as 'r'. Calculate the approximate change in length of the hypotenuse if xx changes to 3 mm and yy changes to 5 mm.

  1. Define the hypotenuse rr and the initial conditions.

    The hypotenuse rr of a right-angled triangle is given by the Pythagorean theorem:

    r=x2+y2r = \sqrt{x^2 + y^2}

    The initial values are x0=2x_0 = 2 mm and y0=3y_0 = 3 mm.
    The changes in xx and yy are:

    Δx=3 mm−2 mm=1 mm\Delta x = 3 \text{ mm} - 2 \text{ mm} = 1 \text{ mm} Δy=5 mm−3 mm=2 mm\Delta y = 5 \text{ mm} - 3 \text{ mm} = 2 \text{ mm}

  2. Find the partial derivatives of rr with respect to xx and yy.

    ∂r∂x=∂∂x(x2+y2)1/2=12(x2+y2)−1/2(2x)=xx2+y2\frac{\partial r}{\partial x} = \frac{\partial}{\partial x}(x^2 + y^2)^{1/2} = \frac{1}{2}(x^2 + y^2)^{-1/2}(2x) = \frac{x}{\sqrt{x^2 + y^2}} ∂r∂y=∂∂y(x2+y2)1/2=12(x2+y2)−1/2(2y)=yx2+y2\frac{\partial r}{\partial y} = \frac{\partial}{\partial y}(x^2 + y^2)^{1/2} = \frac{1}{2}(x^2 + y^2)^{-1/2}(2y) = \frac{y}{\sqrt{x^2 + y^2}}

  3. Evaluate the partial derivatives at the initial values (x0,y0)=(2,3)(x_0, y_0) = (2, 3).

    At (2,3)(2, 3), the initial hypotenuse length is 22+32=4+9=13\sqrt{2^2 + 3^2} = \sqrt{4+9} = \sqrt{13}.

    ∂r∂x∣(2,3)=213\frac{\partial r}{\partial x} \Big|_{(2,3)} = \frac{2}{\sqrt{13}} ∂r∂y∣(2,3)=313\frac{\partial r}{\partial y} \Big|_{(2,3)} = \frac{3}{\sqrt{13}}

  4. Calculate the approximate change in rr using the total differential formula, dr=∂r∂xΔx+∂r∂yΔydr = \frac{\partial r}{\partial x} \Delta x + \frac{\partial r}{\partial y} \Delta y.

    dr=(213)(1)+(313)(2)dr = \left(\frac{2}{\sqrt{13}}\right)(1) + \left(\frac{3}{\sqrt{13}}\right)(2) dr=213+613=813dr = \frac{2}{\sqrt{13}} + \frac{6}{\sqrt{13}} = \frac{8}{\sqrt{13}}

    To rationalize the denominator:

    dr=813⋅1313=81313dr = \frac{8}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}} = \frac{8\sqrt{13}}{13}

    The approximate change in the length of the hypotenuse is:

    81313 mm≈2.22 mm\boxed{\frac{8\sqrt{13}}{13} \text{ mm} \approx 2.22 \text{ mm}}

    1.2 Given that x=t−t2x = t - t^2 and y=ln⁡(3t)y = \ln(3t), determine:

    1.2.1 dydx\frac{dy}{dx}

  5. Find the derivatives of xx and yy with respect to tt.

    dxdt=ddt(t−t2)=1−2t\frac{dx}{dt} = \frac{d}{dt}(t - t^2) = 1 - 2t dydt=ddt(ln⁡(3t))=13t⋅3=1t\frac{dy}{dt} = \frac{d}{dt}(\ln(3t)) = \frac{1}{3t} \cdot 3 = \frac{1}{t}

  6. Use the chain rule for parametric differentiation, dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.

    dydx=1t1−2t\frac{dy}{dx} = \frac{\frac{1}{t}}{1 - 2t} dydx=1t(1−2t)\boxed{\frac{dy}{dx} = \frac{1}{t(1 - 2t)}}

    1.2.2 d2ydx2\frac{d^2 y}{dx^2}

  7. Use the formula for the second derivative of parametric equations

    d2ydx2=ddt(dydx)⋅dtdx\frac{d^2 y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}.
    From 1.2.1, we have dydx=1t(1−2t)=(t−2t2)−1\frac{dy}{dx} = \frac{1}{t(1 - 2t)} = (t - 2t^2)^{-1}.
    Also, dtdx=1dx/dt=11−2t\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{1 - 2t}.

  8. Calculate ddt(dydx)\frac{d}{dt}\left(\frac{dy}{dx}\right).

    ddt((t−2t2)−1)=−1(t−2t2)−2⋅ddt(t−2t2)\frac{d}{dt}\left((t - 2t^2)^{-1}\right) = -1(t - 2t^2)^{-2} \cdot \frac{d}{dt}(t - 2t^2) =−(t−2t2)−2(1−4t)= -(t - 2t^2)^{-2} (1 - 4t) =−1−4t(t(1−2t))2= -\frac{1 - 4t}{(t(1 - 2t))^2}

  9. Multiply the results from Step 1 and Step 2.

    d2ydx2=(−1−4t(t(1−2t))2)⋅(11−2t)\frac{d^2 y}{dx^2} = \left(-\frac{1 - 4t}{(t(1 - 2t))^2}\right) \cdot \left(\frac{1}{1 - 2t}\right) d2ydx2=−1−4tt2(1−2t)3\boxed{\frac{d^2 y}{dx^2} = -\frac{1 - 4t}{t^2(1 - 2t)^3}}
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