Please help me with these GCSE calculus questions.

[nyrychvantel]

1. A curve passes through the point (\pi/2 , 0) and it's gradient at any point (x,y) on the curve is (sin x)(1 - sin x), find the equation of the normal to the curve at the point x = \pi.

I tried to sub x = \pi, then dy/dx = 0
so that means gradient of the tangent at x = \pi is 0
So, can I say the gradient of the normal is also = 0?
However it contradicts me as I know (Grad. of tgt.)(Grad. of normal) = -1
0 * 0 \neq -1
How can I show the equation of the normal to my examiners with mathematical explanations?

2.

Can someone please explain to me why there are two different results(if i ignore the c)? Or did I do the question wrongly?



3. \int\frac{3-x}{1-x}
How can I solve this question? I'm totally stucked because there is a x in the numerator. So is the "ln" method still applicable here?

Thanks!

 

[danago]

For the second one, try the substitution u = 1-x

 

[HallsofIvy]

1. A curve passes through the point (\pi/2 , 0) and it's gradient at any point (x,y) on the curve is (sin x)(1 - sin x), find the equation of the normal to the curve at the point x = \pi.

I tried to sub x = \pi, then dy/dx = 0
so that means gradient of the tangent at x = \pi is 0
So, can I say the gradient of the normal is also = 0?
However it contradicts me as I know (Grad. of tgt.)(Grad. of normal) = -1
0 * 0 \neq -1
How can I show the equation of the normal to my examiners with mathematical explanations?

I presume that by "gradient" of a function of one variable, you mean the derivative. (That's common in England. In the United States, "gradient" is usually reserved for the vector of partial derivatives of a function of several variables.)

No, the "gradient of the normal" is not "also = 0". that ought to be clear if you have any idea what "gradient" means. The gradient is the slope of the tangent line. You want the line perpendicular to that. If the slope of 1 line is 0, then the slope of a line perpendicular to it is -1/m- if it exists. If the slope of a line is 0, it is, of course, horizontal (of the form y= constant). In that case the perpendicular line does not have a "slope" but is of the form x= constant.

2.

Can someone please explain to me why there are two different results(if i ignore the c)? Or did I do the question wrongly?

You can't ignore the c! The only difference between the two is that constant. One of the constants is the other plus ln(3). Since ln(3) is itself a constant, c+ ln(3) is still a constant.

3. \int\frac{3-x}{1-x}
How can I solve this question? I'm totally stucked because there is a x in the numerator. So is the "ln" method still applicable here?

Thanks!

Let u= 1-x. then du= -dx and x= 1- u so 3- x= what?

 

[nyrychvantel]

I presume that by "gradient" of a function of one variable, you mean the derivative. (That's common in England. In the United States, "gradient" is usually reserved for the vector of partial derivatives of a function of several variables.)

No, the "gradient of the normal" is not "also = 0". that ought to be clear if you have any idea what "gradient" means. The gradient is the slope of the tangent line. You want the line perpendicular to that. If the slope of 1 line is 0, then the slope of a line perpendicular to it is -1/m- if it exists. If the slope of a line is 0, it is, of course, horizontal (of the form y= constant). In that case the perpendicular line does not have a "slope" but is of the form x= constant.

Erm...then how do you determine whether the tangent is a horizontal or vertical? I'm not sure I could convince my examiner without any mathematical proof...

You can't ignore the c! The only difference between the two is that constant. One of the constants is the other plus ln(3). Since ln(3) is itself a constant, c+ ln(3) is still a constant.

I get what you mean! thanks!

Let u= 1-x. then du= -dx and x= 1- u so 3- x= what?

I suddenly notice I can break it down by partial fraction.
I've never learned substitution method to solve such question, but I think substitution is better as it can calculate the c straightaway! Please correct me if I do it wrong..