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junior college 2 | H2 Maths
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Pls help w q1&2 pls!!
Date Posted: 6 years ago
Views: 375
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Q1...
Date Posted:
6 years ago
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Q2...
Date Posted:
6 years ago
Why do we have to sub 2 into 1?
I only stop at y=1-x
I only stop at y=1-x
Tangent parallel to x-axis is a perfect horizontal line, y=1-x is not a horizontal line. Must be in the format of y=k where k is a constant.
So this is an unusual case where we have to sub back into the original eqn?
It is unusual if you don't!
dy/dx is just a gradient, you need to form an equation of a straight line for a tangent.
Just that this gradient depends not only on x but also y, make it looks like a straight line eqn.
dy/dx is just a gradient, you need to form an equation of a straight line for a tangent.
Just that this gradient depends not only on x but also y, make it looks like a straight line eqn.
The y=1-x is the result of dy/dx=0, meaning it is the result that cause the gradient to be zero. Not the eqn of the straight line tangent.
Oh okay.. cos Usually when sub dydx =0 will get the eqn of tangent directly, didnt know have to sub back..
Usually after sub dy/dx=0, we found x=constant. This x is the x-coordinate on the curve which has the gradient wanted (in this case, the gradient is zero) then find the corresponding y-coord. Finally we do y-y1=gradient*(x-x1) for the tangent...
This special case gives a linear eqn as the solution to dy/dx, makes us confused that was the eqn of the tangent...
Wish you understand my explanation.
This special case gives a linear eqn as the solution to dy/dx, makes us confused that was the eqn of the tangent...
Wish you understand my explanation.
Ohh alright thank you!
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