Eric Nicholas K's answer to LockB's Secondary 4 A Maths Singapore question.
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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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