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secondary 4 | A Maths
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need help with these qns, pls explain too
Date Posted: 3 years ago
Views: 202
This chapter is too soon to be done at this point considering that you only covered the basic rules of differentiation not long ago.
Nevertheless, Practise Now 1 and Practise Now 2 should be easy.
Nevertheless, Practise Now 1 and Practise Now 2 should be easy.
I just do one question from Practise Now 1.
Q1
A tangent is basically a line drawn to a curve such that it hits the curve around that region once only (though in some cases like cubic graphs, further intersection at a later point is possible)
A normal is a line perpendicular to the tangent at the point of the tangency to the curve. It is not just any random line perpendicular to the tangent - it has to be at the tangential point.
This first section combines differentiation with some coordinate geometry.
y = 2x³ - 9x² + 12x + 7
When x = 3/2, y = 23/2. We need the point (3/2, 23/2) later on.
dy/dx
= 6x² - 18x + 12
= 6 (3/2)² - 18 (3/2) + 12
= -3/2
Note that "dy/dx" represents the gradient of the tangent, and not normal, to the curve at a particular point.
To find the gradient of the normal to the curve at that same point, we just use the fact that the product of the gradients of the two perpendicular lines is -1.
Gradient of tangent = -3/2
Gradient of normal = 2/3
Point that both lines pass through is (3/2, 23/2).
For the tangent,
y = mx + c
23/2 = -3/2 (3/2) + c
23/2 = -9/4 + c
55/4 = c
y = -3/2 x + 55/4
For the normal,
y = mx + c
23/2 = 2/3 (3/2) + c
23/2 = 1 + c
21/2 = c
y = 2/3 x + 21/2
Q1
A tangent is basically a line drawn to a curve such that it hits the curve around that region once only (though in some cases like cubic graphs, further intersection at a later point is possible)
A normal is a line perpendicular to the tangent at the point of the tangency to the curve. It is not just any random line perpendicular to the tangent - it has to be at the tangential point.
This first section combines differentiation with some coordinate geometry.
y = 2x³ - 9x² + 12x + 7
When x = 3/2, y = 23/2. We need the point (3/2, 23/2) later on.
dy/dx
= 6x² - 18x + 12
= 6 (3/2)² - 18 (3/2) + 12
= -3/2
Note that "dy/dx" represents the gradient of the tangent, and not normal, to the curve at a particular point.
To find the gradient of the normal to the curve at that same point, we just use the fact that the product of the gradients of the two perpendicular lines is -1.
Gradient of tangent = -3/2
Gradient of normal = 2/3
Point that both lines pass through is (3/2, 23/2).
For the tangent,
y = mx + c
23/2 = -3/2 (3/2) + c
23/2 = -9/4 + c
55/4 = c
y = -3/2 x + 55/4
For the normal,
y = mx + c
23/2 = 2/3 (3/2) + c
23/2 = 1 + c
21/2 = c
y = 2/3 x + 21/2
ohh.. usually which chapter is taught after the chapter on basic rules of differentiation?
Different books have different order, but this is what I do
- basic rule
- chain rule
- product rule
- quotient rule
- differentiation repeated times
- tangent/normal
- stationary points
- rates of change
- differentiation of trigo
- differentiation of exponential
- differentiation of logarithm
- basic rule
- chain rule
- product rule
- quotient rule
- differentiation repeated times
- tangent/normal
- stationary points
- rates of change
- differentiation of trigo
- differentiation of exponential
- differentiation of logarithm
from my textbook, the chapter after the basic rules of differentiation is this chapter tho (the one i posted here). btw what is the "differentiation repeated times" about?
Basically you can differentiate a function more than once
are you free to help with some questions today? im stucked at some questions on differentiation... (just posted it)
You might see this under the name “higher derivatives”.
I’m unlikely to answer questions today, if tmr I have time I might do
if we want to differentiate an equation like 2t+5+4/5t-9, are we supposed to turn it into a single fraction and apply chain rule or we can just ignore 2t+5 and focus on doing chain rule for 4/5t-9?
We can differentiate them separately as I have done in the other post.
can you help me with the qns in this post too as my teacher is going to start on this chapter soon... sorry for bothering you, really appreciate it
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This is a graphical representation for Practise Now 3. I am lazy to draw out the tangential lines.
Date Posted:
3 years ago
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The first two
Date Posted:
3 years ago
still dont really understand what is normal to the curve...
The tangent to a curve is basically a straight line drawn to hit the curve at one point around that region.
The normal to the curve is basically a straight line, perpendicular to the tangent line, drawn to pass the same tangential point as described above.
The normal to the curve is basically a straight line, perpendicular to the tangent line, drawn to pass the same tangential point as described above.
Oh ya. Forgot to add.
Suppose a line is a tangent to the curve at a point P.
Then the normal to the curve at P is basically the line perpendicular to the tangent and passing the point P.
The normal is NOT just any random perpendicular line to the tangent. It is a specific line.
Suppose a line is a tangent to the curve at a point P.
Then the normal to the curve at P is basically the line perpendicular to the tangent and passing the point P.
The normal is NOT just any random perpendicular line to the tangent. It is a specific line.
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Practise Now 2
The rest I do another day.
The rest I do another day.
Date Posted:
3 years ago
i dont really understand 1(iii)
“The tangent at Q is perpendicular to the tangent at P”
When we talk about perpendicular lines, we always talk about gradients. Remember that tangents are lines themselves.
At P, the gradient of the tangential line is 1. By using the fact that the gradients of P and Q multiply to -1, we find that the gradient of the tangent at Q must be -1.
But you know that dy/dx represents the gradient of the curve at a specific point.
So, dy/dx = -1 at Q.
From there, you solve the equation accordingly for Q.
4x - 7 = -1
And so on.
When we talk about perpendicular lines, we always talk about gradients. Remember that tangents are lines themselves.
At P, the gradient of the tangential line is 1. By using the fact that the gradients of P and Q multiply to -1, we find that the gradient of the tangent at Q must be -1.
But you know that dy/dx represents the gradient of the curve at a specific point.
So, dy/dx = -1 at Q.
From there, you solve the equation accordingly for Q.
4x - 7 = -1
And so on.
sorry!! still dont really understand why is it 4x-7=-1 and why it can give us the answer
why is the gradient same as the normal to the curve of P but they are 2 different lines tho (this qn is really confusing for me)
why is the gradient same as the normal to the curve of P but they are 2 different lines tho (this qn is really confusing for me)
Remember that dy/dx = 4x - 7 from part i?
Recall that dy/dx represents the gradient of the tangent to the curve at a point. Of course, with different values of x, you will get a different value of dy/dx, indicating a changing gradient with a changing value of x.
At Q, dy/dx has a particular value. This value is -1, simply because
"...the tangent at point Q is perpendicular to the tangent at point P..."
This statement came from the question, one line above the question.
You must see this question from the coordinate geometry point of view. Because we found from part (i) that the gradient of the tangent line at P is 1, it follows that ANY line perpendicular to P must have a gradient of -1 (because -1 x 1 = -1). This includes the required tangent at Q, which is why the gradient of the curve at Q must be -1 based on these pieces of information.
At Q, dy/dx = 4x - 7 for some unknown value of x. But we just found that at Q, the gradient of the tangent is -1. The dy/dx and the gradient represent the exact same thing, which is why we can say that at Q, dy/dx = -1, so
4x - 7 = -1
As for the normal thing, hang on for a moment while I put the sketch on Desmos.
Recall that dy/dx represents the gradient of the tangent to the curve at a point. Of course, with different values of x, you will get a different value of dy/dx, indicating a changing gradient with a changing value of x.
At Q, dy/dx has a particular value. This value is -1, simply because
"...the tangent at point Q is perpendicular to the tangent at point P..."
This statement came from the question, one line above the question.
You must see this question from the coordinate geometry point of view. Because we found from part (i) that the gradient of the tangent line at P is 1, it follows that ANY line perpendicular to P must have a gradient of -1 (because -1 x 1 = -1). This includes the required tangent at Q, which is why the gradient of the curve at Q must be -1 based on these pieces of information.
At Q, dy/dx = 4x - 7 for some unknown value of x. But we just found that at Q, the gradient of the tangent is -1. The dy/dx and the gradient represent the exact same thing, which is why we can say that at Q, dy/dx = -1, so
4x - 7 = -1
As for the normal thing, hang on for a moment while I put the sketch on Desmos.
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LockB, this is a supplement to the Practise Now 2 Q1.
DIfferent points on the curve y = 2x² - 7x will have different gradients and therefore different equations of tangents.
Now, we know that points P and Q will produce clearly different tangential lines, indicated by the blue and red lines in the picture (they do not appear perpendicular to the scaling which I have used, but they are actually perpendicular if drawn to equal scales on both axes).
This is because it is given in the question that the tangent to the curve at Q is perpendicular to the tangent to the curve at P. In short, we simply draw two tangents from two different points so that they meet at right angles.
Note that the tangent to the point Q is not the same as the normal to the curve at P because the tangent to the point Q does not pass point P at all. The normal to the curve at P would need to be perpendicular to the tangent at P and pass through point P (as indicated by the purple line).
--------------------------------------------------------------
So, the purple and red lines are obviously parallel to each other, because we know that
- the normal to the curve at P is, by default, perpendicular to the tangent to the curve at P
- the tangent to the curve at Q is perpendicular to the tangent to the curve at P (given in the question).
But, as explained previously, these are not the same.
When we talk about a normal to the curve, we are strictly referring to that one perpendicular line passing through the tangential point only. Other lines will never be identical to the normal.
The tangent to the curve at point Q, which happens to be perpendicular to the tangent to the curve at P, can just be treated as "any other perpendicular line" when you are looking at point P.
DIfferent points on the curve y = 2x² - 7x will have different gradients and therefore different equations of tangents.
Now, we know that points P and Q will produce clearly different tangential lines, indicated by the blue and red lines in the picture (they do not appear perpendicular to the scaling which I have used, but they are actually perpendicular if drawn to equal scales on both axes).
This is because it is given in the question that the tangent to the curve at Q is perpendicular to the tangent to the curve at P. In short, we simply draw two tangents from two different points so that they meet at right angles.
Note that the tangent to the point Q is not the same as the normal to the curve at P because the tangent to the point Q does not pass point P at all. The normal to the curve at P would need to be perpendicular to the tangent at P and pass through point P (as indicated by the purple line).
--------------------------------------------------------------
So, the purple and red lines are obviously parallel to each other, because we know that
- the normal to the curve at P is, by default, perpendicular to the tangent to the curve at P
- the tangent to the curve at Q is perpendicular to the tangent to the curve at P (given in the question).
But, as explained previously, these are not the same.
When we talk about a normal to the curve, we are strictly referring to that one perpendicular line passing through the tangential point only. Other lines will never be identical to the normal.
The tangent to the curve at point Q, which happens to be perpendicular to the tangent to the curve at P, can just be treated as "any other perpendicular line" when you are looking at point P.
Date Posted:
3 years ago
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This is an example of a normal to the curve - look at the three-way intersection between the curve, the tangent and the normal.
Date Posted:
3 years ago
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This is an example of a scenario where two different random tangents drawn to the curve happens to meet each other at right angles at some point not lying on the curve.
Remember that many tangent lines can be drawn to the curve, and these two lines are just some of them. There is no three-way intersection at a common point this time, so this is not considered a normal to the curve.
However, it is a fact that this line is parallel to the normal line at the other point mentioned in the previous example.
Remember that many tangent lines can be drawn to the curve, and these two lines are just some of them. There is no three-way intersection at a common point this time, so this is not considered a normal to the curve.
However, it is a fact that this line is parallel to the normal line at the other point mentioned in the previous example.
Date Posted:
3 years ago
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Practise Now 3
Date Posted:
3 years ago
don't really understand why do we find the two gradients and why do we equate them together to find the coordinates....
don't really understand why do we find the two gradients and why do we equate them together to find the coordinates....
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In this case, we are defining the gradient of the exact same required line in two ways.
One, by considering the fact that the line is a tangent to the curve - it has a gradient (given by the expression for dy/dx) and it passes through the point (x, y).
Two, by considering the fact that we can find the gradient of the line by simply taking the gradient between two points on the line (x, y) and (1, 4). This is using the gradient formula y2 - y1 divided by x2 - x1.
Of course, you are talking about the exact same line with the same gradient, so we must equate them to solve for the value of x.
--------------------------------------------------------
In this case, we are defining the gradient of the exact same required line in two ways.
One, by considering the fact that the line is a tangent to the curve - it has a gradient (given by the expression for dy/dx) and it passes through the point (x, y).
Two, by considering the fact that we can find the gradient of the line by simply taking the gradient between two points on the line (x, y) and (1, 4). This is using the gradient formula y2 - y1 divided by x2 - x1.
Of course, you are talking about the exact same line with the same gradient, so we must equate them to solve for the value of x.
A lot of these questions which you have seen requires you to bring together many different concepts from different chapters together. And some of them, as is the case for this question, will require you to form equations to solve for x.
When you learn algebraic expressions for the first time, you first learnt how to substitute the value of x to find the value of y. This skill trains you to obtain an output value given an input value, and is clearly staightforward. You would have seen this in Primary 6, assuming you attended the Primary 6 education in Singapore.
Eventually, you are asked to find the value of x when you are provided with the value of y. This skill trains you to form equations to solve for the input value which can give us the output value. You would have seen this in Secondary 1.
These are the same thing, but finding the input value (in Sec 1) is much harder than finding the output value (in Pri 6), simply because it works that way,
Here in differentiation, it's the same, First, you are asked to find the value of dy/dx when provided with the value of x.
Afterwards, like the case here, you are asked to find the value of x which leads to a certain expression. Being able to formulate equations to solve for an unknown x is very critical in A Maths.
When you learn algebraic expressions for the first time, you first learnt how to substitute the value of x to find the value of y. This skill trains you to obtain an output value given an input value, and is clearly staightforward. You would have seen this in Primary 6, assuming you attended the Primary 6 education in Singapore.
Eventually, you are asked to find the value of x when you are provided with the value of y. This skill trains you to form equations to solve for the input value which can give us the output value. You would have seen this in Secondary 1.
These are the same thing, but finding the input value (in Sec 1) is much harder than finding the output value (in Pri 6), simply because it works that way,
Here in differentiation, it's the same, First, you are asked to find the value of dy/dx when provided with the value of x.
Afterwards, like the case here, you are asked to find the value of x which leads to a certain expression. Being able to formulate equations to solve for an unknown x is very critical in A Maths.
we can use one point on the curve to find the coordinates of 2 different tangent points? sorry this question is really confusing me...
dont really understand the part on "let a point on the curve...." and "then the gradient of the tangent formed...."
"we can use one point on the curve to find the coordinates of 2 different tangent points? sorry this question is really confusing me..."
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For this question, the graph is actually a little unusual, in the sense that there are actually TWO possible points on the curve which can be drawn together with the point (1, 4) separately in two different straight lines to form two entirely different tangent lines which can satisfy the question.
---------------------------------------------------------
"dont really understand the part on "let a point on the curve...." and "then the gradient of the tangent formed....""
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You need to define a random point (x, y) to be connected to the point (1, 4) with a straight line so that you can find the gradient of the straight line using the gradient formula (y2 - y1) / (x2 - x1).
"The gradient of the line formed" is simply this formula being used here.
After using the formula, you get gradient = (y - 4) / (x - 1).
But then, this point is not just any random point (x, y) on the graph. It must be a point lying on the curve y = x / (x + 1). So, the y-coordinate of the curve must of course be x / (x + 1) for any given value of x. This is why the coordinate becomes what I wrote.
---------------------------------------------------------
For this question, the graph is actually a little unusual, in the sense that there are actually TWO possible points on the curve which can be drawn together with the point (1, 4) separately in two different straight lines to form two entirely different tangent lines which can satisfy the question.
---------------------------------------------------------
"dont really understand the part on "let a point on the curve...." and "then the gradient of the tangent formed....""
--------------------------------------------------------
You need to define a random point (x, y) to be connected to the point (1, 4) with a straight line so that you can find the gradient of the straight line using the gradient formula (y2 - y1) / (x2 - x1).
"The gradient of the line formed" is simply this formula being used here.
After using the formula, you get gradient = (y - 4) / (x - 1).
But then, this point is not just any random point (x, y) on the graph. It must be a point lying on the curve y = x / (x + 1). So, the y-coordinate of the curve must of course be x / (x + 1) for any given value of x. This is why the coordinate becomes what I wrote.
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Practise Now 4, Practise Now 5
Date Posted:
3 years ago
rate of change is basically dy/dx?
dont really understand practice now 5 tho...
dont really understand practice now 5 tho...
Rate of change is basically the study of how one variable changes as another variable changes.
dy/dx is basically a measure of how y changes as x changes for a particular curve (which is again yet another way of expressing the word gradient). Thus, dy/dx can be called the rate of change of y with respect to x.
When the rate of change involves time, we are simply measuring how fast a quantity is changing with time. For example, we know that speed = distance / time. We can say that speed is a measure of how fast the distance changes, so speed can be said to be the rate of change of distance (s) with time (t); this can be written as ds/dt.
In Practise Now 5, we are dealing with even more difficult scenarios. Connected rates of change. In many circumstances, the change of a variable can affect yet another variable.
For example, when we inflate a balloon, it is not just the radius which changes. The surface area and the volume also changes! We wish to find, say, how fast the volume changes when the radius changes constantly. And so on. We wish to know how changing the radius affects the change in volume - will it always be a constant increase? Will the balloon volume increase faster if the radius gets larger? And so on.
dy/dx is basically a measure of how y changes as x changes for a particular curve (which is again yet another way of expressing the word gradient). Thus, dy/dx can be called the rate of change of y with respect to x.
When the rate of change involves time, we are simply measuring how fast a quantity is changing with time. For example, we know that speed = distance / time. We can say that speed is a measure of how fast the distance changes, so speed can be said to be the rate of change of distance (s) with time (t); this can be written as ds/dt.
In Practise Now 5, we are dealing with even more difficult scenarios. Connected rates of change. In many circumstances, the change of a variable can affect yet another variable.
For example, when we inflate a balloon, it is not just the radius which changes. The surface area and the volume also changes! We wish to find, say, how fast the volume changes when the radius changes constantly. And so on. We wish to know how changing the radius affects the change in volume - will it always be a constant increase? Will the balloon volume increase faster if the radius gets larger? And so on.
In this question, we wish to find out how the surface area of a sphere changes as its radius increases constantly.
The rate of change of surface area is called dA/dt. The rate of change of the radius is called dr/dt.
These two are clearly related expressions, because the surface area of a balloon increases every time a balloon is inflated to a greater radius.
But how do we link them together? We link these two using a connector expression dA/dr using the equation
dA/dt = dA/dr times dr/dt.
We are given a constant, fixed value for dr/dt. We are tasked to find the value of dA/dt. To do this, we need to connect them using dA/dr. But, this must be obtained from an equation for A in terms of r. This is easy because we know the formula for the curved surface area of a sphere.
From there, we can write an expression for dA/dr before proceeding to solve the question.
The rate of change of surface area is called dA/dt. The rate of change of the radius is called dr/dt.
These two are clearly related expressions, because the surface area of a balloon increases every time a balloon is inflated to a greater radius.
But how do we link them together? We link these two using a connector expression dA/dr using the equation
dA/dt = dA/dr times dr/dt.
We are given a constant, fixed value for dr/dt. We are tasked to find the value of dA/dt. To do this, we need to connect them using dA/dr. But, this must be obtained from an equation for A in terms of r. This is easy because we know the formula for the curved surface area of a sphere.
From there, we can write an expression for dA/dr before proceeding to solve the question.
is the rate of change and rate of increase different? as i dont understand why the answer isnt 96pi and 57pi (like practice now 4)
still dont really understand the part on "the rate of increase of the surface area and volume....."
still dont really understand the part on "the rate of increase of the surface area and volume....."
Rate of change and rate of increase is similar, just like “percentage change” versus “percentage increase”.
I look at the rates of change again later.
I look at the rates of change again later.
ok
"the rate of increase of the surface area and volume....."
Basically we are required to obtain an expression for dA/dt and dV/dt.
Unfortunately, we do not have these values directly, so we must make use of other expressions to obtain our required expressions.
dA/dt can be found by multiplying dA/dr by dr/dt, both quantities of which can be found easily because we already know the value of dr/dt and we can easily calculate the value of dA/dr when r = 12, considering that the area mentioned here is the surface area of a sphere.
dV/dt is similar and we use dV/dr times dr/dt.
Basically we are required to obtain an expression for dA/dt and dV/dt.
Unfortunately, we do not have these values directly, so we must make use of other expressions to obtain our required expressions.
dA/dt can be found by multiplying dA/dr by dr/dt, both quantities of which can be found easily because we already know the value of dr/dt and we can easily calculate the value of dA/dr when r = 12, considering that the area mentioned here is the surface area of a sphere.
dV/dt is similar and we use dV/dr times dr/dt.
hi Mr Eric, although this question is from 3months ago, is the rate of increase (dont really remember if it is increase or decrease) always negative regardless of the signage of the answer you obtained?
for example i got 0.886cm^3/s as my answer, but the question is asking for rate of increase, so i will have to add a negative sign to make it - 0.886cm^3/s
for example i got 0.886cm^3/s as my answer, but the question is asking for rate of increase, so i will have to add a negative sign to make it - 0.886cm^3/s
When we calculate dV/dt or something like that, it’s always a change with respect to the positive. So if the answer is -0.886, then it means the increase is -0.886, or basically this is a decrease of 0.886.
sorry! i still dont really understand when to add a negative sign to an answer....
by the way are you able to help with some questions on integration of partial fractions?
by the way are you able to help with some questions on integration of partial fractions?
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