Draw a tangent that is perpendicular to the line

[TSN79]

I have a curve -\sqrt {2x^3 } on which I'm supposed to draw a tangent that is perpendicular to the line \[y = \frac{4}{3}x + \frac{1}{3}.

I know that this tangent must have a "steepness" of -3/4 in order to make it perpendicular, but how do I now find the point on the graph?

 

[EnumaElish]

Solve for the x from the curve formula that would make its slope equal to that of the tangent.

 

[Diane_]

Given that the tangent needs to be perpendicular to the line y = 4/3x + 1/3, he's right that the slope must be -3/4. Negative reciprocal for the normal line.

TSN79 - What you need is the point on the curve where the first derivative is -3/4. Just take the derivative, plug in that value, and solve for x.

Note: I'm presuming you're taking or have taken calculus, as I don't see any other way to do this.

 

[EnumaElish]

Note: I'm presuming you're taking or have taken calculus, as I don't see any other way to do this.

He could do it geometrically I suppose, but that may or may not be exact.

 

[TSN79]

Just take the derivative, plug in that value, and solve for x.

Take the derivative of what? And plug in what value? -3/4? I tried with <br /> - \sqrt {2x^3 } but didn't really get anywhere...

 

[EnumaElish]

Take the derivative of - \sqrt {2x^3 } then equate it to -3/4 and solve for x.

 

[TSN79]

Ah, now where getting somewhere, thanks you all!

 

[HallsofIvy]

Did you really need to be told that? You were told that the line had to be tangent to the curve given by y=-\sqrt {2x^3 }. Didn't you connect "tangent" with "derivative"- after you had been told the find the derivative- the only tangent mentioned was to y= -\sqrt {2x^3 }.
.