Question

Implicit Differential
I need help with part B and C please
suppose that sq. rt. of x + sq. rt. of y = 10
a) find dy/dx. I answered it which is y'=-(y 1/2) / (x 1/2)
b) what is the slope of the tangent to this curve which passes through (36,16)? (Confirm first that the curve does in fact, pass through this point)
c) find an equation for this line

Answer 1

\(\huge \sqrt{36}+\sqrt{16}=10 \) CONFIRMED....

Answer 2

(36, 16) is on the implicit curve.

Answer 3

your derivative is correct so just plug in those x, y values into your derivative...

Answer 4

Thank you~
and for part C do I just take 6 and 4 as part of the equation?

Answer 5

where did 6, 4 come from? use (36, 16) as the point.
your slope comes from the derivative...
once you have the slope, you can write the equation of the tangent line in point-slope form.

Answer 6

oooh I got it from the sq. rt. that you showed me on how to confirm it
so, after I plugged in the x and y values it becomes -8/5 which is the slope right? (:

Answer 7

point slope from: \(\large y-y_0=m(x-x_0) \)
where m is your slope, and \(\large (x_0,y_0) \) is the point

Answer 8

y-16=-8/5(x-36)

Answer 9

not quite... your slope is wrong....

Answer 10

slope is not -8/5

Answer 11

your derivative is correct: \(\large y=-\frac{y^{1/2}}{x^1/2}=-\frac{\sqrt y}{\sqrt x}=-\frac{\sqrt{36}}{\sqrt{16}}=-\frac{3}{2} \)

Answer 12

so your slope should be -3/2

Answer 13

My mistake was that I plugged it into (y1/2)/(x1/2) with the 1/2
I understand now. Thank you so much