Find the equation of the straight line that is tangent to th

[kimmy_koo51]

I actually have a few questions. I am not too sure how to enter the square root symbol so I will just write it out

1. Find the equation of the straight line that is tangent to the graph g(x)+ -3x^2 where the tangent line is perpendicular to the line 3y+2x-15=0.

2. Fine the values of x so that the tangent to the function 3/3(square root)(x^2) is parallel to the line x+16y+3=0

3. A motorboat coasts toward a dock with its engine off. Its distance s, in meters, from the dock t seconds after the engine is turned off is s(t)= 10(6-t)/t+3 for 0 is equal to or < t is equal to or < 6.
a. How far is the boat from the dock initially?
b. Find the boat's velocity when it bumps into the dock (No it is not stoped)

Thank you very much in advance.

 

[skeeter]

Re: Find the equation of the straight line that is tangent t

I actually have a few questions. I am not too sure how to enter the square root symbol so I will just write it out

1. Find the equation of the straight line that is tangent to the graph g(x)+ -3x^2 where the tangent line is perpendicular to the line 3y+2x-15=0.


1. find the value of the slope perpendicular to the given line.
2. set the value of the perpendicular slope found in step 1 equal to g'(x) and solve for x.
3. using the value of x found in step 2, calculate the y-value of the point on using g(x).
4. you now have a point, (x,y) and a slope ... use the point-slope form of a linear equation to determine the requisite equation.

2. Fine the values of x so that the tangent to the function 3/3(square root)(x^2) is parallel to the line x+16y+3=0

... uhhh, 3/[3sqrt(x^2)] = 1/|x| ... do you really mean
\(\displaystyle \L \frac{3}{\sqrt[3]{x^2}}\) ???

3. A motorboat coasts toward a dock with its engine off. Its distance s, in meters, from the dock t seconds after the engine is turned off is s(t)= 10(6-t)/t+3 for 0 is equal to or < t is equal to or < 6.
a. How far is the boat from the dock initially?
b. Find the boat's velocity when it bumps into the dock (No it is not stoped)

same problem ...http://www.freemathhelp.com/forum/viewtopic.php?t=18148

Thank you very much in advance.

 

[kimmy_koo51]

Hey...thanks a lot, and yes that's what I meant in the second question if you could still help me out.

 

[skeeter]

2. Fine the values of x so that the tangent to the function 3/3(square root)(x^2) is parallel to the line x+16y+3=0

1. find the slope of the given line

2. find the derivative of your function \(\displaystyle \L y = \frac{3}{\sqrt[3]{x^2}} = 3x^{-\frac{2}{3}}\)

3. set derivative = slope of the line found in step 1 ... solve for x.

 

[kimmy_koo51]

For that question I get to when you make them equal eachother but I am not too sure how to solve for x. I have:

-2/x^5/3 = -1/16

 

[skeeter]

\(\displaystyle \L -\frac{2}{x^{\frac{5}{3}}} = -\frac{1}{16}\)

cross multiply ...

\(\displaystyle \L x^{\frac{5}{3}} = 32\)

\(\displaystyle \L (x^{\frac{5}{3}})^{\frac{3}{5}} = 32^{\frac{3}{5}}\)

can you finish now?