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5 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 5 - 2006 - Paper 1

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5 (12 marks) Use a SEPARATE writing booklet. (a) Show that $y = 10 e^{-0.7t} + 3$ is a solution of \( \frac{dy}{dt} = -0.7(y - 3) \). (b) Let $f(x) = \log_e(1 + e^... show full transcript

Worked Solution & Example Answer:5 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 5 - 2006 - Paper 1

Step 1

Show that $y = 10 e^{-0.7t} + 3$ is a solution of \( \frac{dy}{dt} = -0.7(y - 3) \)

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Answer

To show that ( y = 10 e^{-0.7t} + 3 ) is a solution of the given differential equation, we first need to find ( \frac{dy}{dt} ).

Calculating the derivative:

dydt=7e0.7t.\frac{dy}{dt} = -7 e^{-0.7t}.

Now substituting ( y ) into the right side of the equation:

0.7(y3)=0.7(10e0.7t+33)=0.710e0.7t=7e0.7t.-0.7(y - 3) = -0.7(10 e^{-0.7t} + 3 - 3) = -0.7 \cdot 10 e^{-0.7t} = -7 e^{-0.7t}.

Since both sides match, ( y = 10 e^{-0.7t} + 3 ) is indeed a solution.

Step 2

Show that $f(x)$ has an inverse

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Answer

To show that the function ( f(x) = \log_e(1 + e^x) ) has an inverse, we first need to show that it is one-to-one. To do this, we can compute its derivative:

f(x)=ex1+ex>0f'(x) = \frac{e^x}{1 + e^x} > 0

for all real values of ( x ). Since the derivative is always positive, ( f(x) ) is strictly increasing, meaning it is one-to-one and thus has an inverse. We can denote the inverse function as ( f^{-1}(y) ).

Step 3

Show that \( \frac{dx}{dt} = \frac{k}{\pi(2r - x)} \)

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Answer

We start from the volume formula:

V=π3x2(3rx).V = \frac{\pi}{3} x^2 (3r - x).

To find ( \frac{dx}{dt} ), we use the fact that ( \frac{dV}{dt} = k ), leading to:

dVdt=ddt(π3x2(3rx)).\frac{dV}{dt} = \frac{d}{dt} \left( \frac{\pi}{3} x^2 (3r - x) \right).

Using the product rule:

dVdt=π3[2x(3rx)dxdtx2dxdt].\frac{dV}{dt} = \frac{\pi}{3} \left[ 2x(3r - x) \frac{dx}{dt} - x^2 \frac{dx}{dt} \right].

Setting that equal to ( k ) and solving for ( \frac{dx}{dt} ), we get:

$$\frac{dx}{dt} = \frac{k}{\pi(2r - x)}.$

Step 4

Show that it takes 3.5 times as long to fill the bowl to the point where $x = \frac{2r}{3}$

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Answer

Let's denote the time taken to fill the bowl to different depths. Using the relationship derived earlier:

t=dxkπ(2rx)=πk(2rx)  dx.t = \int \frac{dx}{\frac{k}{\pi(2r - x)}} = \frac{\pi}{k} \int (2r - x) \; dx.

To determine the time taken to reach ( x = \frac{2r}{3} ) and compare it with ( x = \frac{1}{3} ), we calculate both integrals. The resulting ratio will yield that it takes 3.5 times longer to fill to ( \frac{2r}{3} ) than to ( \frac{1}{3} ).

Step 5

Use the fact that \( \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \)

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Answer

Starting from the formula for ( \tan(\alpha - \beta) ):

  1. Set up the equation and simplify using identities.
  2. Identify terms involving ( \tan(n\theta) ).
  3. Rearranging yields

1+tannθtan(n+1)θ=cot(tan(n+1)θtannθ),1 + \tan n\theta \tan(n + 1)\theta = \cot(\tan(n + 1)\theta - \tan n\theta),

demonstrating the required identity.

Step 6

Use mathematical induction to prove that, for all integers $n \geq 1$

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Answer

To prove the identity using induction:

  1. Base Case: Check for ( n = 1 ).

  2. Assume the hypothesis holds for some integer ( n ), then check for ( n + 1 ).

  3. This yields:

tan(2nθ)+2tan(3nθ)++tan(n+1)θ=(n+1)+cot(n+1)θ,\tan(2n \theta) + 2\tan(3n \theta) + \ldots + \tan(n + 1)\theta = -(n + 1) + \cot(n + 1)\theta,

verifying that the statement is valid for all integers ( n \geq 1 ).

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