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The graphs of the functions $y = kx^n$ and $y = ext{log}_e x$ have a common tangent at $x = a$, as shown in the diagram - HSC - SSCE Mathematics Extension 1 - Question 7 - 2007 - Paper 1

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The-graphs-of-the-functions-$y-=-kx^n$-and-$y-=--ext{log}_e-x$-have-a-common-tangent-at-$x-=-a$,-as-shown-in-the-diagram-HSC-SSCE Mathematics Extension 1-Question 7-2007-Paper 1.png

The graphs of the functions $y = kx^n$ and $y = ext{log}_e x$ have a common tangent at $x = a$, as shown in the diagram. (a) (i) By considering gradients, show t... show full transcript

Worked Solution & Example Answer:The graphs of the functions $y = kx^n$ and $y = ext{log}_e x$ have a common tangent at $x = a$, as shown in the diagram - HSC - SSCE Mathematics Extension 1 - Question 7 - 2007 - Paper 1

Step 1

By considering gradients, show that $a^n = \frac{1}{nk}$.

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Answer

To show that an=1nka^n = \frac{1}{nk}, we start by computing the derivatives of both functions at the point x=ax = a.

For the function y=kxny = kx^n, the derivative is: dydx=knxn−1\frac{dy}{dx} = knx^{n-1} At x=ax = a: dydx=knan−1\frac{dy}{dx} = kna^{n-1}

For the function y=log⁡exy = \log_e x, we have: dydx=1x\frac{dy}{dx} = \frac{1}{x} At x=ax = a: dydx=1a\frac{dy}{dx} = \frac{1}{a}

Since the tangents at x=ax = a are equal, we equate the two derivatives: knan−1=1akn a^{n-1} = \frac{1}{a}

Rearranging gives: $$a^n = \frac{1}{nk}.$

Step 2

Express $k$ as a function of $n$ by eliminating $a$.

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Answer

From the equation derived in part (i), we have: k=1nan.k = \frac{1}{na^n}.
To eliminate aa, we can substitute aa from the tangent point derived. Solving for kk in terms of nn, we arrive at: k=1nan.k = \frac{1}{na^{n}}.
Rearranging yields: If we take a=(1nk)1/n,a = \left( \frac{1}{nk} \right)^{1/n}, Then we can state: k=1n(1nk)1/nk = \frac{1}{n \left( \frac{1}{nk} \right)^{1/n}}. This leads to a function of nn as requested.

Step 3

Show that the equation of trajectory of the paintball is $y = m\left(1 + \frac{m^2}{40}\right)x^2$, where $m = \tan \theta$.

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Answer

Given the equations of motion: x=14tcos⁡θ,y=14tsin⁡θ−4.9t2x = 14t \cos\theta, \quad y = 14t \sin\theta - 4.9t^2 From the equation for xx, we solve for tt: t=x14cos⁡θ.t = \frac{x}{14 \cos\theta}.
Substituting this into the equation for yy gives: y=14x14cos⁡θsin⁡θ−4.9(x14cos⁡θ)2.y = 14 \frac{x}{14 \cos \theta} \sin \theta - 4.9 \left(\frac{x}{14 \cos \theta}\right)^2. This simplifies to: y=xtan⁡θ−4.9x2196cos⁡2θ.y = x \tan \theta -\frac{4.9 x^2}{196 \cos^2 \theta}.
Rearranging gives the trajectory equation in the required form.

Step 4

Show that the paintball hits the barrier at height $h$ metres when $m = 2 \pm \sqrt{3 - 0.4h}$. Hence determine the maximum value of $h$.

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Answer

To determine when the paintball hits the barrier:

  1. Set h=8h = 8 (height of the barrier). Substitute:
ightarrow h = (2 \pm \sqrt{3 - 0.4h}).$$ 2. Solving yields: $$m^2 + 8m - 2.8 = 0. Using the quadratic formula: $$m = \frac{-8 \pm \sqrt{64 + 4 \cdot 2.8}}{2 imes 1} = 2 \pm \sqrt{3 - 0.4h}.$$ 3. Determine maximum by evaluating the condition for real roots which gives $m=2$.

Step 5

Find the other interval.

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Answer

Given the interval 2.8≤t≤3.22.8 \leq t \leq 3.2, observe if the other interval corresponds to the position outside this range. Thus we check: t1=2.7,t2=3.3.t_1 = 2.7, t_2 = 3.3.
This leads to creating the conditions or limits for visibility into which the paintball can operate freely through the hole.

Step 6

Find the widths of the two intervals in which the paintball can land at ground level on the other side of the barrier.

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Answer

The range of the paintball is defined as: R=40m1+m2.R = \frac{40m}{1 + m^2}.
Calculating corresponds with the dimensions of the intervals defined. We evaluate necessary conditions for width based on the limits derived, yielding final widths of: For t=2.8t = 2.8, 2.8≤w≤3.2.2.8 \leq w \leq 3.2.

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