# Thread: Differentiation - Tangents and normals - 4 diff. problems

1. ## Differentiation - Tangents and normals - 4 diff. problems

i have about 4 problems here all together

my first problem is this question letter H , below is my working out but my answer seems to be wrong so i dont know what im doing wrong.
Find the equation of (i) the tangent and (ii) the normal to the curve.

my working

i) 17x - y + 27 = 0
ii) x + 17y + 121 = 0
the other 3 problems are

3: x = 1/2
4: 87'53'
5: x = 1/3/4

in my text book it doesnt really give me examples directly related to this question to solve it so im quite clueless on question 3 , 4 and 5. The other questions in this chapter are primarily based on questions like 4 and 5. So if someone can help me with those questions i'll be able to do the rest.

thx all help is appreciated.

2. Hello,

There is an error in the derivative.

$(\frac{u}{v})'=\frac{u'v - uv'}{v^2}$

And you put +

3. Originally Posted by KavX
i have about 4 problems here all together

my first problem is this question letter H , below is my working out but my answer seems to be wrong so i dont know what im doing wrong.
Find the equation of (i) the tangent and (ii) the normal to the curve.

my working

i) 17x - y + 27 = 0
ii) x + 17y + 121 = 0
[snip]
Thank you thank you thank you. If ever the virtue of a student showing their work was needed to be seen, here it is!

I can tell you very easily the mistake you've made. The quotient rule you're using is wrong. There's meant to be a minus, NOT a plus, in the numerator of the quotient rule formula. Your derivative and hence gradient are wrong for this reason.

4. Originally Posted by KavX
[snip]the other 3 problems are

3: x = 1/2
4: 87'53'
5: x = 1/3/4

in my text book it doesnt really give me examples directly related to this question to solve it so im quite clueless on question 3 , 4 and 5. The other questions in this chapter are primarily based on questions like 4 and 5. So if someone can help me with those questions i'll be able to do the rest.

thx all help is appreciated.
For 3. and 4. you need to use the fact that $m = \tan \theta$ where m is the gradient of the line and $\theta$ is the angle the line makes with the positive direction of the x-axis.

And for 3., 4. and 5. you do know that the gradient of a tangent is given by dy/dx, right?

Some hints:

3. Solve $\frac{dy}{dx} = \tan 45^0$ for x and $\frac{dy}{dx} = \tan 135^0$ for x .....

4. First find the value of m = dy/dx at x = 2.

5. Solve $\frac{dy}{dx} = \frac{1}{2}$ for x.

5. thanx such a stupid mistake i made, think i got to use to using the product rule in the previous chapter since it has the +.
I can tell you very easily the mistake you've made. The quotient rule you're using is wrong. There's meant to be a minus, NOT a plus, in the numerator of the quotient rule formula. Your derivative and hence gradient are wrong for this reason.
just need help on the last 3 now.

thx again.

6. Originally Posted by KavX
thanx such a stupid mistake i made, think i got to use to using the product rule in the previous chapter since it has the +.

just need help on the last 3 now.

thx again.
See post #4. If, after having a good think, you're still stuck just say so - preferably showing your work again.

7. For the 5th question, the relation Mr F wrote is due to the fact that if 2 lines are parallel, their slope is equal.

The slope a of a line is defined by : y=ax+b

Here, you have x-2y-1=0 <-> y=x/2 -1/2

-> the slope is 1/2

8. so for Q3

y = x^ 2

y' = 2x

so since tan 45 = 1

2x = 1

so x = 1/2

is this how your meant to solve it?

for Q4

my working

oh i c so,

tan @ = 27

so inverse tan 27 = 87 ' 52 ' 44.05

sweet

still dont understand question 5 at all

thanx for the other 2.

9. Did you understand why you were asked to solve dy/dx=tan(45°) or tan(135°) ? :-)

For the 5th question, also remember that :

The value of the derivate at a point is the slope of the tangent at this point. This is why you compare the value of the derivate with the slope of the line you want to be parallel with the tangent.

I don't know if it's clear enough :s

10. dont worry about 5 just got it now lol. thanx for all the help

11. no i dont really understand why i had to use tan,
apart from the fact that the topic is like about tangents unless thats just the reason

heres my working for question 5

12. Originally Posted by KavX
dont worry about 5 just got it now lol. thanx for all the help
And question 4 ....... you do realise you also have to solve dy/dx = tan(135)?

That's because the question doesn't specify whether you want 45 degrees from the positive direction or 45 degrees from the negative direction (which is the same as 135 degrees from the positive direction).

13. So for the tan.

As mentioned, the value of the derivative is the slope of the tangent.

The slope of a line going through points A and B is also defined as : $\frac{y_B-y_A}{x_B-x_A}$, which is the tan if you take the point on the x-axis.

14. Originally Posted by KavX
no i dont really understand why i had to use tan,
apart from the fact that the topic is like about tangents unless thats just the reason

heres my working for question 5

15. Originally Posted by mr fantastic
And question 4 ....... you do realise you also have to solve dy/dx = tan(135)?

That's because the question doesn't specify whether you want 45 degrees from the positive direction or 45 degrees from the negative direction (which is the same as 135 degrees from the positive direction).

so technically what your saying there should be 2 answers for this question

x = 1/2 and x = -1/2 ??

Page 1 of 2 12 Last