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Serggg [28]
3 years ago
7

What is the equation linethatis perpendicular to the given line and passes through the point (3,0)?​

Mathematics
1 answer:
vladimir1956 [14]3 years ago
3 0

Answer:

3y+x = 3

Step-by-step explanation:

for the point ( 3,0)

x=3 , y= 0

the equation for the line is given as

y-y1= m(x-x1)

for perpendicular ,

m= -1/x

so

y-0 = -1/3(x-3)

3y= -x +3

3y+x=3

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3 years ago
B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =
jolli1 [7]

I suppose you mean

g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)

Differentiate one term at a time.

Rewrite the first term as

\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}

Then the product rule says

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'

Then with the power and chain rules,

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}

Simplify this a bit by factoring out \frac12 (36-x^2)^{-3/2} :

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}

For the second term, recall that

\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}

Then by the chain rule,

\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}

So we have

g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}

and we can simplify this by factoring out 18(36-x^2)^{-3/2} to end up with

g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}

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