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July 30, 2021, 05:31:37 am

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#### fun_jirachi

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##### Re: VCE Methods Question Thread!
« Reply #19155 on: June 14, 2021, 02:18:48 pm »
+1

The answer in the solutions is correct - can you show us your working, so we can see where you went wrong? It's difficult to help you out without much context.

The area is just the area of the trapezium minus the definite integral over the same domain. You seem to have gotten only the area of the trapezium wrong.
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#### aspiringantelope

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##### Re: VCE Methods Question Thread!
« Reply #19156 on: June 14, 2021, 03:37:45 pm »
0

The answer in the solutions is correct - can you show us your working, so we can see where you went wrong? It's difficult to help you out without much context.

The area is just the area of the trapezium minus the definite integral over the same domain. You seem to have gotten only the area of the trapezium wrong.
Apologies
The attached image (which is the alternative way of solving the question and the method I chose) does not equate to the answer provided in the solution?
Unless I have excluded the trapezium?
Regards
« Last Edit: June 14, 2021, 03:40:43 pm by aspiringantelope »

#### fun_jirachi

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##### Re: VCE Methods Question Thread!
« Reply #19157 on: June 14, 2021, 04:31:42 pm »
+8
You're calculating the area of a larger trapezium that does not match the one in the question.

The correct integral is $\int_0^{5\ln 4} \frac{-3x}{10\ln 4} + 2 - 2e^{-\frac{x}{5}}$. If you split up the integral into the integral of the curve and the integral for the trapezium it should become abundantly clear that at $x = 5\ln 4$ the corresponding y-value is not 1/2 for the function that defines the top of the trapezium, but rather 7/2. This is why you're returning a value much larger than the correct answer - you haven't excluded the trapezium but you've calculated an additional area on top of the trapezium.

Hope this helps
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#### Corey King

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##### Re: VCE Methods Question Thread!
« Reply #19158 on: June 14, 2021, 05:22:50 pm »
0
https://snipboard.io/px8q0n.jpg

https://snipboard.io/6ykVvc.jpg

https://snipboard.io/2rtf8j.jpg

Hey guys,

Does anyone happen to know how to solve these questions?
I've given them my best shot, but clearly don't know how to solve them.

Many thanks,
Corey

#### fun_jirachi

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##### Re: VCE Methods Question Thread!
« Reply #19159 on: June 14, 2021, 05:53:29 pm »
+8
The function you've found in part a) is parabolic - find the value of x for which f(x) is a minimum and substitute back into the function. You can do this by using properties of a parabola (the vertex is always at a fixed x value given the coefficients of the parabola) or by using calculus (differentiate and find where the derivative is zero).
The last part of the question involves the chain rule: note that $\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}$. You're given $\frac{dV}{dt}$, and you can easily calculate $\frac{dV}{dx}$ by differentiating the function from part a). Isolate $\frac{dx}{dt}$ and substitute in x = 6 appropriately to get the answer.

If the function is piecewise continuous, $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x)$. Find values of p and q such that px+q has the same value at x = 5 as the second piece. You've done this part correctly.

If the function is piecewise differentiable, $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x)$ and $\lim_{x \to 5^-} f'(x) = \lim_{x \to 5^+} f'(x)$. Find values for p and q that satisfy the first equation in part a) but this time, you can find the exact values for p and q as f'(5) must be equal for both pieces.

You can then find the corresponding range of the function given both pieces.

Note your derivative for the first piece is wrong, it should be p

part b) is wrong. I'm not quite sure how you ascertained that the tangent does not cross the coordinate axes. Draw out the graph again, it's just the hyperbola $y = \frac{1}{x}$ for $x < 0$. It definitely does. You can use the point gradient form of the line to find the intercepts. Given the intercepts, it should be easy enough to get the area of the triangle OPQ as the area is just $\frac{OP \times OQ}{2}$. This should result in a constant, which is independent of m.
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#### parieeelol

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##### Re: VCE Methods Question Thread!
« Reply #19160 on: June 14, 2021, 06:59:51 pm »
0
Hey guys, just came across this question in the Cambridge textbook and am completely stumped on how the worked solutions derived the length of the triangle for part ii)

My coordinates for part i) were (3+√1/a , 0)  and (3−√1/a, 0) and I thought I could get the length by finding the distance between the points, and solved in on my CAS but got a completely different answer. I've attached the worked solutions for the question as well.

Would appreciate any help, thank you

#### fun_jirachi

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##### Re: VCE Methods Question Thread!
« Reply #19161 on: June 14, 2021, 07:31:07 pm »
+7
If B is the point $\left(3 + \sqrt{\frac{1}{a}}, 0\right)$, A is the point $\left(3 - \sqrt{\frac{1}{a}}, 0\right)$, then $|BA| = 2\sqrt{\frac{1}{a}}$. What was the answer you got? I'm still unsure as to exactly what part you're stuck on

Recall that for any two points $s, t$ with coordinates $(x_1, y_1)$ and $(x_2, y_2)$ respectively we have that the distance $|st| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. Here, we just have $|BA| = \left|\left(3 + \sqrt{\frac{1}{a}}\right) - \left(3 - \sqrt{\frac{1}{a}}\right)\right| = 2\sqrt{\frac{1}{a}}$
« Last Edit: June 14, 2021, 07:36:47 pm by fun_jirachi »
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#### Corey King

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##### Re: VCE Methods Question Thread!
« Reply #19162 on: June 14, 2021, 08:30:46 pm »
0
The function you've found in part a) is parabolic - find the value of x for which f(x) is a minimum and substitute back into the function. You can do this by using properties of a parabola (the vertex is always at a fixed x value given the coefficients of the parabola) or by using calculus (differentiate and find where the derivative is zero).
The last part of the question involves the chain rule: note that $\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}$. You're given $\frac{dV}{dt}$, and you can easily calculate $\frac{dV}{dx}$ by differentiating the function from part a). Isolate $\frac{dx}{dt}$ and substitute in x = 6 appropriately to get the answer.

If the function is piecewise continuous, $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x)$. Find values of p and q such that px+q has the same value at x = 5 as the second piece. You've done this part correctly.

If the function is piecewise differentiable, $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x)$ and $\lim_{x \to 5^-} f'(x) = \lim_{x \to 5^+} f'(x)$. Find values for p and q that satisfy the first equation in part a) but this time, you can find the exact values for p and q as f'(5) must be equal for both pieces.

You can then find the corresponding range of the function given both pieces.

Note your derivative for the first piece is wrong, it should be p

part b) is wrong. I'm not quite sure how you ascertained that the tangent does not cross the coordinate axes. Draw out the graph again, it's just the hyperbola $y = \frac{1}{x}$ for $x < 0$. It definitely does. You can use the point gradient form of the line to find the intercepts. Given the intercepts, it should be easy enough to get the area of the triangle OPQ as the area is just $\frac{OP \times OQ}{2}$. This should result in a constant, which is independent of m.

Thanks Jirachi, I now understand the first two links. But the third link I'm still lost on. I have no idea how to get those intercept points. How can I find a tangent line anyways that isn't in terms of m or x? Is m meant to represent the gradient or something? I don't know

#### fun_jirachi

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##### Re: VCE Methods Question Thread!
« Reply #19163 on: June 14, 2021, 08:54:36 pm »
+6
You've found that the gradient is $-\frac{1}{m^2}$ at $x = -m$. This implies that the tangent line at $x = -m$ is $y + \frac{1}{m} = -\frac{1}{m}(x+m)$. This is because you also have the coordinates $(-m, -\frac{1}{m})$. I'm not sure what you mean by 'How can I find a tangent line anyways that isn't in terms of m or x?' -  you don't have to. The area of the triangle isn't in terms of m or x, but the tangent line is. You use the tangent line to find the intercepts and hence the area bounded by the coordinate axes and the tangent line. I'm not sure how you drew that conclusion :/.

Hope this is clearer now
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#### Corey King

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##### Re: VCE Methods Question Thread!
« Reply #19164 on: June 14, 2021, 09:19:49 pm »
0
You've found that the gradient is $-\frac{1}{m^2}$ at $x = -m$. This implies that the tangent line at $x = -m$ is $y + \frac{1}{m} = -\frac{1}{m}(x+m)$. This is because you also have the coordinates $(-m, -\frac{1}{m})$. I'm not sure what you mean by 'How can I find a tangent line anyways that isn't in terms of m or x?' -  you don't have to. The area of the triangle isn't in terms of m or x, but the tangent line is. You use the tangent line to find the intercepts and hence the area bounded by the coordinate axes and the tangent line. I'm not sure how you drew that conclusion :/.

Hope this is clearer now

I can see how we have that x coordinate, and that I know the gradient at point -m. But I don't see how you get the y-intercept form values? (here: $x = -m$ is $y + \frac{1}{m} = -\frac{1}{m}(x+m)$ )
In my head, if -m = 0 we have our y intercept. But if -m = 0, then 1/m^2 is undefined. So there would be an asymptote or something? I don't yet know how to extract those values as you did.

#### fun_jirachi

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##### Re: VCE Methods Question Thread!
« Reply #19165 on: June 14, 2021, 09:30:16 pm »
+9
Note that we are calculating the tangent to the curve $y = \frac{1}{x}$ for $x < 0$. This is the first important piece of information that you're missing - $m$ is strictly non-zero. This means the tangent line is always defined. Note also that $m$ is just a constant, you should be treating it like one.

The second thing you're missing here is that $m = 0$ does not mean you have a y-intercept. You may have one, but a y-intercept occurs on the Cartesian plane if and only if $x = 0$.

The last piece that I can clearly see you're not understanding is the values I plugged into the point-gradient form of the line. I didn't pull those values out of thin air. If our curve is $y = \frac{1}{x}, \ x < 0$ and we have an arbitrary point at $x = -m$, then we have a point on the curve $\left(-m, -\frac{1}{m}\right)$. I trust you've seen the point-gradient form of the line $y - y_0 = m(x-x_0)$, where $m$ is the gradient of the line, and $(x_0, y_0)$ is a point on the line. I just plugged the point $\left(-m, -\frac{1}{m}\right)$ and the gradient at $x = -m$ which you calculated in the first part into the formula. Note that the gradient at $x = -m$ is the gradient of the curve, and hence this line is the tangent at that point. $m$ can never be zero. I didn't extract anything.

Hope this is completely clear this time
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#### biology1234

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##### Re: VCE Methods Question Thread!
« Reply #19166 on: June 15, 2021, 10:45:17 pm »
0
The line that passes through (-4,3) and (5,-2 ) also passes through :
A   (7,0)
B (1/2, 1/2)
C  (-14,-7)
D  (-5,2)
E ( 13,-8)

How would you guys work this one out. Would appreciate

#### biology1234

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##### Re: VCE Methods Question Thread!
« Reply #19167 on: June 15, 2021, 10:49:09 pm »
0
Find the equation of the line perpendicular to 3x- 2y+6 =0 and having the same y-intercept.
What would be the process would greatly appreciate.
Thank you

#### fun_jirachi

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##### Re: VCE Methods Question Thread!
« Reply #19168 on: June 15, 2021, 11:21:39 pm »
+8
Please avoid double-posting and instead amend your previous post using the modify button. It would also help to show some evidence of previous work (working out, what you know about the question, what you have tried). Just some tips for future questions

If you have two points on the xy-plane, you can construct a unique line on the xy-plane. Given the two points $(x_1, y_1)$ and $(x_2, y_2)$ on the line, the line has equation $y-y_1 = \frac{y_2 - y_1}{x_2-x_1} (x-x_1)$ - (why does this work? that's something for you to consider, this will help you understand more questions in future). Here, we're given the two points $(-4, 3)$ and $(5, -2)$. Construct the line using the directions I've given, and see which points satisfy the equation; the one point that does satisfy the equation will be the answer.

The y-intercept of a line is calculated by letting $x = 0$. After you find this point, we must also find the gradient of the line provided. Here, the gradient is $\frac{3}{2}$ (I didn't pull that number out of thin air; how did I get it? Another question for you to consider). recall that perpendicular lines on the xy-plane have the product of their gradients equalling -1 ie. the equation of the line perpendicular to the line provided will have gradient $-\frac{2}{3}$. We can construct a unique line given the gradient of a line and a point it passes through. This point is the y-intercept you would have found earlier. Use the point-gradient form of the line from the previous question to find the equation of this particular line.

Hope this helps

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#### arnavg2207

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##### Re: VCE Methods Question Thread!
« Reply #19169 on: June 27, 2021, 03:32:04 pm »
0
Hi!
I needed help with this normal distribution question (Ex 16D Q5).
The weights of cats are normally distributed. It is known that 10% of cats weigh more
than 1.8 kg, and 15% of cats weigh less than 1.35 kg. Find the mean and the standard
deviation of this distribution.

I can find the probability for normal distributions, and can use the inverse to find the value for that probability, but am struggling with questions that ask you to find the mean/standard deviation.  I also need help with this question, but perhaps if I learn how to solve the previous one I can solve this one as well :-).

Machine A is packaging bags of mints with a mean weight of 300 grams. The bags
are considered underweight if they weigh less than 295 grams. It is observed that, on
average, 5% of bags are rejected as underweight. Assuming that the weights of the
bags are normally distributed, find the standard deviation of the distribution.
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