need help with a tangent and normal question (1 Viewer)

wgy182 · Jul 11, 2013

Can anyone help me with this question?

Find the gradient of the tangent on the curve y=x/sqrt (2x) at the point (4, sqrt(2)). Express your answer as a rational number.

Thanks in advance

Makematics · Jul 11, 2013

simply the equation and differentiate?

wgy182 · Jul 11, 2013

I did but I keep getting the wrong answer

Makematics · Jul 11, 2013

hmmm well i get 1/4root2, although this may well be incorrect as i rushed it. i'll type it up for you if it is correct.

wgy182 · Jul 11, 2013

Ok thanks but the answer in the book is m= sqrt(2)-2/2

HeroicPandas · Jul 11, 2013

Makematics said:
hmmm well i get 1/4root2, although this may well be incorrect as i rushed it. i'll type it up for you if it is correct.

Meanwhile, in makematics's mind:

$y = \frac{x}{\sqrt{2x} }\\ \\y = \frac{x}{\sqrt{2}\sqrt{x}} \\ y = \frac{\sqrt{x}}{\sqrt{2}} \\ \frac{dy}{dx} = \frac{1}{2}\cdot \frac{1}{\sqrt{2}\sqrt{x}} \\ \frac{dy}{dx} \left ( x = 4 \right ) = \frac{1}{2}\cdot \frac{1}{2\sqrt{2}} \\ \\~~~~~~~~~~~~~~~= \frac{\sqrt{2}}{8}$

Makematics · Jul 11, 2013

wat? 'express your answer as a rational number'... is dis possible?

JT145 · Jul 12, 2013

@Heroic Pandas

In your third line did you multiply the whole thing by the square root of x

Because then wouldn't the numerator be x (square root of x)

If that makes sense

Menomaths · Jul 12, 2013

JT145 said:
@Heroic Pandas

In your third line did you multiply the whole thing by the square root of x

Because then wouldn't the numerator be x (square root of x)

If that makes sense

When you multiply by square root x you get x square root x on the numerator but you also get a x at the denominator. The denominator x cancel's the numerator x leaving you with square root x / square root 2.

girlworld_club · Jul 12, 2013

lol. how did i manage to get 8 / root 2. exactly the reverse.

how would you work out this answer without simplifying it first. if anyone is bothered that would be much appreciated.

HeroicPandas · Jul 12, 2013

girlworld_club said:
lol. how did i manage to get 8 / root 2. exactly the reverse

feel free to post up ur working out, and any BoSer will check it and identify and correct mistakes

girlworld_club · Jul 12, 2013

This is going to be so hard to see:

Ok.

y = x/root2x
= x(2x)^1/2
y' = (2x)^1/2 + 1/2(2x)^-1/2 . 2 . x
= root 2x + x/ root 2x
sub 4 in:
= root 8 + 4/root8
= 12/ root8
= 3root 8/2

what did i do wrong??

HeroicPandas · Jul 12, 2013

girlworld_club said:
This is going to be so hard to see:

Ok.

y = x/root2x
= x(2x)^1/2
y' = (2x)^1/2 + 1/2(2x)^-1/2 . 2 . x
= root 2x + x/ root 2x
sub 4 in:
= root 8 + 4/root8
= 12/ root8
= 3root 8/2

what did i do wrong??

bringing index up, means pwoer is negative

girlworld_club · Jul 12, 2013

wooopsss. but i still don't manage to get the correct answer. now it just complicates things by having

y' = 1/ root2x - x / root (2x)^3

bedpotato · Jul 12, 2013

girlworld_club said:
wooopsss. but i still don't manage to get the correct answer. now it just complicates things by having

y' = 1/ root2x - x / root (2x)^3

Try using the quotient rule. That's what I used without simplifying it, and I got sqrt(2)/8.

HeroicPandas · Jul 12, 2013

girlworld_club said:
wooopsss. but i still don't manage to get the correct answer. now it just complicates things by having

y' = 1/ root2x - x / root (2x)^3

such a complicated pathway would only provide u misery, so we must look for another way to approach this question

The golden rule for all differentiating or integrating question: simplify the question so that it can be easily differentiated or integrated

MENTALITY (A)

$y = \frac{x}{\sqrt{2x}} \\ \\ y = \frac{x}{\sqrt{2}\cdot \sqrt{x}} \left ( $square root laws: \sqrt{ab} = \sqrt{a}\cdot\sqrt{b} \right ) \\ \\ y = $\frac{x^{1-\frac{1}{2}}}{\sqrt{2}} \left ( $using index laws: $\frac{x^a}{x^b} = x^{a-b} \right ) \\ \\ y = \frac{x^{-\frac{1}{2}}}{\sqrt{2}} \\ \\ y = \frac{1}{\sqrt{2}}\cdot x^{\frac{1}{2}}$

MENTALITY (B)

$y = \frac{x}{\sqrt{2x}} \\ \\ y = x\cdot(2x)^{-\frac{1}{2}} \\ \\ y = x\cdot (2)^{-\frac{1}{2}}(x)^{-\frac{1}{2}} \left ( $index laws: (ab)^m = a^m\cdot b^m \right ) \\ \\ y = \frac{x^{1+\left ( \frac{-1}{2} \right )}}{\sqrt{2}} \left ( $ $index laws: x^a\cdot x^b = x^{a+b} \right ) \\ \\ y = $\frac{x^{\frac{1}{2}}}{\sqrt{2}}$

This is attractive function can be easily differentiated

girlworld_club · Jul 12, 2013

thanks mate, i'll keep that in mind before diving into differentiating next time. however, i just don't understand why my method doesn't work. it's go tto be possible somehow.

ymcaec · Jul 12, 2013

girlworld_club said:
wooopsss. but i still don't manage to get the correct answer. now it just complicates things by having

y' = 1/ root2x - x / root (2x)^3

you will still get 1/4sqrt2 if u sub x = 4

girlworld_club · Jul 12, 2013

ymcaec said:
you will still get 1/4sqrt2 if u sub x = 4

rly..am i subbing wrong?

Menomaths · Jul 12, 2013

HeroicPandas said:
such a complicated pathway would only provide u misery, so we must look for another way to approach this question

The golden rule for all differentiating or integrating question: simplify the question so that it can be easily differentiated or integrated

MENTALITY (A)

$y = \frac{x}{\sqrt{2x}} \\ \\ y = \frac{x}{\sqrt{2}\cdot \sqrt{x}} \left ( $square root laws: \sqrt{ab} = \sqrt{a}\cdot\sqrt{b} \right ) \\ \\ y = $\frac{x^{1-\frac{1}{2}}}{\sqrt{2}} \left ( $using index laws: $\frac{x^a}{x^b} = x^{a-b} \right ) \\ \\ y = \frac{x^{-\frac{1}{2}}}{\sqrt{2}} \\ \\ y = \frac{1}{\sqrt{2}}\cdot x^{-\frac{1}{2}}$

MENTALITY (B)

$y = \frac{x}{\sqrt{2x}} \\ \\ y = x\cdot(2x)^{-\frac{1}{2}} \\ \\ y = x\cdot (2)^{-\frac{1}{2}}(x)^{-\frac{1}{2}} \left ( $index laws: (ab)^m = a^m\cdot b^m \right ) \\ \\ y = \frac{x^{1+\left ( \frac{-1}{2} \right )}}{\sqrt{2}} \left ( $ $index laws: x^a\cdot x^b = x^{a+b} \right ) \\ \\ y = $\frac{x^{-\frac{1}{2}}}{\sqrt{2}}$

This is attractive function can be easily differentiated

For mentality A, line 3 shouldn't it be x^1+1/2? Isn't the power of 1/square root x = -1/2?

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