Question

Find the slope of the tangent to the curve y = 3x^4 - 4x at x = 4

And

Find the equation of tangent to the curve given by

x = a sin^3 t, y = bcos^3 t at a point where t = pi/2

Answer 1

Hmm I was trying to figure out that second one...

So like....The slope of the tangent line is given by \(\Large\rm \dfrac{dy}{dx}\).

We have parametric equations so for our dy/dx we have,\[\Large\rm \frac{dy}{dx}=\frac{(dx/dt)}{(dy/dt)}\]yes?

Answer 2

Woops I think I wrote that upside down. :P

Answer 3

\[\Large\rm \frac{dy}{dx}=\frac{(dy/dt)}{(dx/dt)}\]Ya ya ya, my bad.

Answer 4

Take your derivatives, simplify it all down and you should get something like this,
\[\Large\rm \frac{dy}{dt}=-\frac{b}{a}\cot(t)\]Then evaluate it at \(\Large\rm t=\frac{\pi}{2}\) to get the `slope` of your tangent line.

We're trying to form a line: \(\Large\rm y=mx+b\).

Answer 5

I think you can handle those derivatives in the middle, yes? c:
Those shouldn't give you too much trouble.
Just a little chain rule going on.

How bout the first problem you posted?
Stuck on that one also?

Answer 6

If u messed up with derivatives..(as i am too)..i know how to do it algebraically..

when x= 4
for tangent slope...(instantaneous) you plug in the very next or before number like

if it asked for x=4
u can plug in either x=3.9999
or
x=4.0001

if it has asked for x=1(for example)
then for the second value u can either plug in x=0.999 or x=1.001

now when you have the values of y1 and y2

use this formula


slope (m)=(y2 - y1)/(x2-x1)

that 'l give u your answer!


hope this helps!