# Math Help - Questions in derivatives

1. ## Questions in derivatives

2. Originally Posted by r-soy

To #1a):

The equation of the line is:

$y = -\frac1{24} x + 1$

Therefore the perpendicular direction to this line has the slope m = 24.

Therefore you have to solve for x:

$f'(x)=24$

Remark: Probably you have solved $f'(x)=0$

3. Originally Posted by r-soy

To 1.b):

The given line has the slope m = -12.

Therefore you have to solve for x:

f'(x) = -12

To get the y-coordinaste of the points in question you plug in the x-coordinates into the term of f.

4. my dear help me about Q 1 for a how we got m = 24 and for b = - 12

plese write the step .

5. for 1 a how we got m = 24 and for b = - 12 ?
and what is the point for a and b ?

2 ) are 2 and 3 the answre correct now:

6. Originally Posted by r-soy
my dear help me about Q 1 for a how we got m = 24 and for b = - 12

plese write the step .
Hopefully, the math gurus will help me, but I THINK that, since you're looking for the line that is perpendicular to $y = 1 - (\frac{x}{24})$, the slope of the line you're looking for will be the negative reciprocal of the slope of the line you have. So, for the equation of the line $y = 1 - (\frac{x}{24})$, you have that the slope is:

$-(\frac{x}{24}) = (-\frac{1}{24})x \Longrightarrow -\frac{1}{24}$

Now, the negative of this slope is $-(-\frac{1}{24}) \Longrightarrow \frac{1}{24}$

The reciprocal, then, is 24.

Again, let those who know correct me... ;-)

7. For (1)(a) the problem asked where the tangent line was perpendicular to y= 1- x/24. You found x where the derivative was 0. I have no idea why you did that! The derivative is the slope of the tangent line. A line with slope 0 is horizontal and is perpendicular to a vertical line. y= 1- x/24 is NOT vertical. It has slope -1/24. Clearly you are expected to know that if y= mx+ b and y= nx+ d are perpendicular, then n= -1/m. A line perpendicular to a line with slope -1/24 must have slope 24. THAT was why everyone set the derivative equal to 24, not 0!

For (1)(b), the problem asked where the tangent line was parallel to $y= \sqrt{3}- 12x$ which has slope -12. Two lines are parallel if and only if they have the same slope.

Again- two lines, y= mx+ b and y= nx+ c, are parallel if and only if m= n, perpendicular if and only if m= -1/n (equivalently mn= -1).

8. thanls a lot plese .. see My participat Number 6

9. is correct :