Find the value of k

BabyMilo · #1 $Old$ November 15th, 2009, 01:05 PM

The point P on the curve $y=k\sqrt{x}$ has x-coordinate 4. the normal to the curve at P is parallel to the line 2x+3x=0

what's the value of k.

many thanks!

Soroban · #2 $Old$ November 15th, 2009, 02:02 PM

Hello, BabyMilo!

Quote:

The point $P$ on the curve: $y\:=\:k\sqrt{x}$ has $x$ -coordinate 4.

The normal to the curve at $P$ is parallel to the line: $2x+3x\:=\:0$

What's the value of $k$ ?

The graph has the equation: . $y \:=\:kx^{\frac{1}{2}}$

The tangent to the graph has slope: . $y' \:=\:k\cdot\frac{1}{2}x^{-\frac{1}{2}} \:=\:\frac{k}{2\sqrt{x}}$

At $P\:(x = 4)$ , the slope of the tangent is: . $m_t \:=\:\frac{k}{2\sqrt{4}} \:=\:\frac{k}{4}$

Hence, at $P$ , the slope of the normal is: . $m_n \:=\:\frac{4}{k}$

The line $2x + 3y \:=\:0 \quad\Rightarrow\quad y \:=\:-\frac{2}{3}x$ .has slope $-\tfrac{2}{3}$

Hnce: . $\frac{4}{k} \:=\:-\frac{2}{3} \quad\Rightarrow\quad\boxed{ k \:=\:-6}$

BabyMilo · #3 $Old$ November 15th, 2009, 02:31 PM

ok thanks for your help

but can i just check

Quote:

The tangent to the graph has slope: . $y' \:=\:k\cdot\frac{1}{2}x^{-\frac{1}{2}} \:=\:\frac{k}{2\sqrt{x}}$

should be $y' = k*\frac{1}{2}x^{-\frac{1}{2}} = \frac{\frac{1}{2}k}{2\sqrt{x}}$

Quote:

At $P\:(x = 4)$ , the slope of the tangent is: . $m_t \:=\:\frac{k}{2\sqrt{4}} \:=\:\frac{k}{4}$

At $P(x = 4)$ , the slope of the tangent is: . $\frac{\frac{1}{2}k}{2\sqrt{4}} = \frac{\frac{1}{2}k}{2}$

Quote:

Hence, at $P$ , the slope of the normal is: . $m_n \:=\:\frac{4}{k}$

Hence, at $P$ ,

the slope of the normal is: . $\frac{-2}{\frac{1}{2}k}$

Quote:

Hnce: . $\frac{4}{k} \:=\:-\frac{2}{3} \quad\Rightarrow\quad\boxed{ k \:=\:-6}$

Hnce: . $\frac{-2}{\frac{1}{2}k} = -\frac{2}{3} \quad\Rightarrow\quad\boxed{ k =+6}$

ukorov · #4 $Old$ November 15th, 2009, 11:07 PM

Quote:

Originally Posted by Soroban $View Post$

The graph has the equation: . $y \:=\:kx^{\frac{1}{2}}$

The tangent to the graph has slope: . $y' \:=\:k\cdot\frac{1}{2}x^{-\frac{1}{2}} \:=\:\frac{k}{2\sqrt{x}}$

At $P\:(x = 4)$ , the slope of the tangent is: . $m_t \:=\:\frac{k}{2\sqrt{4}} \:=\:\frac{k}{4}$

Hence, at $P$ , the slope of the normal is: . $m_n \:=\:\frac{4}{k}$

The line $2x + 3y \:=\:0 \quad\Rightarrow\quad y \:=\:-\frac{2}{3}x$ .has slope $-\tfrac{2}{3}$

Hnce: . $\frac{4}{k} \:=\:-\frac{2}{3} \quad\Rightarrow\quad\boxed{ k \:=\:-6}$

can not quite understand why the slope of tangent = $\frac{k}{2 \sqrt{x}}$ ...?

another thing: when slope of tangent is $\frac{k}{4}$ , the slope of normal is $-\frac{4}{k}$ , isn't it?