#dy over dx
37 messages · Page 1 of 1 (latest)
Ohh thankss
I forgot to use the straight line eqatuon
And stopped after derfintion
for second one you pretty much do the same, find the gradient of the tangent through differentiation, however as its the normal it is perpendicular to the tangent so you use the negative reciprocal of the gradient of the tangent
Thankss
np
I hVe a q
Why is 4x^3-1
And not 4x^3
because -x differentiates to -1
Ohh I thought like -1 was joined onto the power
as its -x^1, so multiply by power, and then reduce the power by 1, so -1^0
ahh nah its seperate
Emm what's the 2nd to last line
Oh wait I think I get it
so for the equation of the line you need gradient and a point it passes through
the question tells you it passes through x=1, so you can sub x=1 into the original equation to find the y value at that point, so you have a point it passes through
How do u do the last one
Like I got x + 5y = 33
so differentiate and you get 2x-1 right?
Yeah
and since the normal is perpendicular to the tangent use the negative reciprocal of the tangent as the gradient of the normal, and it passes through (3,6)
so y-6=-1/5(x-3)
rearrange for y=mx+c by first multiplying by 5, so 5y-30=-x+3
add x to both sides and 30 to both sides, so 5y+x=33
and when it touches x axis is when y=0
so just sub y=0
so x=33
so the point is (33,0)
Thanks 🌚