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Mathematics 16 Online
OpenStudy (czesc):

INSTANT FAN AND MEDAL EASY For what value of x is the slope of the curve undefined for the graph defined by x = 10 - t^2, y = t^3 - 12t?

OpenStudy (solomonzelman):

The same thing again, \(\color{#000000}{ \displaystyle \frac{dy}{dx}=\frac{ \displaystyle \frac{dy}{dt} }{ \displaystyle \frac{dx}{dt}} }\)

OpenStudy (solomonzelman):

Alternatively, \(\color{#000000}{ \displaystyle y'(x)={\rm slope}=\frac{ \displaystyle y'(t) }{ \displaystyle x'(t)} }\)

OpenStudy (solomonzelman):

and as again, you are then to see when is this expression for slope is undefined; for what values of t?

OpenStudy (czesc):

i still have no idea what to do

OpenStudy (phi):

do you know how to find the derivative wrt to "t" of x = 10 - t^2 ?

OpenStudy (phi):

*with respect to "t"

OpenStudy (czesc):

Yeah I derived it, but I'm not sure what to do after deriving @phi , thanks for responding. x'=-2t {+c} y'=3t^2-12 {+c}

OpenStudy (phi):

yes, by x' means dx/dt and y' means dy/dt slope of a curve with point (x,y) is defined to by dy/dx

OpenStudy (phi):

slope of a curve with points (x,y) is defined to be dy/dx

OpenStudy (phi):

in other words, notice that dy/dt * dt/dx = dy/dx or if you like \[ \frac{ \frac{dy}{dt} }{\frac{dx}{dt}}= \frac{dy}{dx}= slope \]

OpenStudy (czesc):

Ah, that makes a lot of sense, thank you so much!

OpenStudy (phi):

btw, there is no additional +c when you take derivatives that only happens when you integrate

OpenStudy (phi):

thus the slope is \[slope= \frac{3t^2 -12}{-2t} \]

OpenStudy (czesc):

-3x^2+12 / 2x , and this would be undefined at x=0 correct i mean t not x oops

OpenStudy (phi):

and it will be undefined when t= 0

OpenStudy (czesc):

Thanks a lot!

OpenStudy (phi):

yes, exactly

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