Please help!
1) Find the point on the curve $$y=x \sqrt{x}$$ where the tangent line is parallel to the line 6x-y=4
2) At what point on the curve y=-2x^4 is the tangent line is perpendicular to the line x-y+1=0.

Question

Please help!
1) Find the point on the curve $$y=x \sqrt{x}$$ where the tangent line is parallel to the line 6x-y=4
2) At what point on the curve y=-2x^4 is the tangent line is perpendicular to the line x-y+1=0.

anonymous · Answer

i got the the first the question!

anonymous · Answer

what does it mean by perpendicular?

anonymous · Answer

@pooja195

ganeshie8 · Answer

Heyy still working on this ?

anonymous · Answer

yes

anonymous · Answer

@ganeshie8

anonymous · Answer

right now, i got y'=-8x^3

mathmath333 · Answer

$\large \begin{align} \color{black}{\normalsize \text{the slope of the line  6x-y=4 is}\hspace{.33em}\~\
y=6x-4  \hspace{.33em}\~\
m=6 \hspace{.33em}\~\
\normalsize \text{ For our tangent line to be parallel it must have the same slope. }\hspace{.33em}\~\
\normalsize \text{  So when is the slope of the tangent line 6 }\hspace{.33em}\~\
 \normalsize \text{ Set the derivate equal to 6 }\hspace{.33em}\~\
y=x\sqrt{x} \hspace{.33em}\~\
y=x^{\frac{3}{2}} \hspace{.33em}\~\
y'=\dfrac{3}{2}x^{\frac{1}{2}} \hspace{.33em}\~\
6=\dfrac{3}{2}x^{\frac{1}{2}} \hspace{.33em}\~\
x^{\frac{1}{2}}=4  \hspace{.33em}\~\
x=16  \hspace{.33em}\~\
\normalsize \text{But to find a point, we need x and  y value.   }\hspace{.33em}\~\
\normalsize \text{ So plug 16 in for your x value into your original equation   }\hspace{.33em}\~\
y=6x-4  \hspace{.33em}\~\
y=6\times 16-4  \hspace{.33em}\~\
y=92  \hspace{.33em}\~\
\LARGE (x,y)\implies (16,92)  \hspace{.33em}\~\

}\end{align}$

anonymous · Answer

i think you sub in the wrong equation @mathmath333

anonymous · Answer

need help on the second question please. @ganeshie8