If z = f(x, y), where f is differentiable, and the following applies, find dz/dt when t = 1.?

Question

Given that ( z = f(x, y) ), where ( f ) is a differentiable function, please find ( \frac{dz}{dt} ) when ( t = 1 ). The relationships are defined as follows: ( x = g(t) ) and ( y = h(t) ), with the specific values ( g(1) = -5 ), ( h(1) = 1 ), ( g'(1) = -6 ), and ( h'(1) = 1 ). Additionally, the partial derivatives at the point are ( f_x(-5, 1) = 1 ) and ( f_y(-5, 1) = -1 ).

Anonymous · Accepted Answer

laughing is correct up til"dz/dt = -6  1 + 1  -1dz/dt = -6 &ndash; 2dz/dt = -8"It&rsquo;s easy to fix though (just 1 * -1 = -1, not -2):dz/dt = -6  1 + 1  -1dz/dt = -6 &ndash; 1dz/dt = -7

Anonymous · Answer

If z = f(x,y)(dz/dt) = (dx/dt)  (dz/dx) + (dy/dt)  (dz/dy) Where dz/dx and dz/dy are partial derivatives. I am not sure what fx(-5,1) = 1 and fy(-5,1) = -1 mean, but I guess they represent these partial derivatives. Summarizing the information until now:g(1) = -5 &ndash;> x = -5 in this special case for t = 1h(1) = 1 &ndash;> y = 1 in this special case for t = 1g'(1) = -6h'(1) = 1for those specific values of x and y, the partial derivatives are:dz/dx = 1dz/dy = -1Filling in gives:(dz/dt) = (dx/dt)  (dz/dx) + (dy/dt)  (dz/dy)dz/dt   =     -6         1     +     1          -1dz/dt   =           -6                        &ndash; 2 dz/dt = -8Edit: I have to admit, that&rsquo;s a bit stupid :). Thanks for correcting me!

Anonymous

If z = f(x, y), where f is differentiable, and the following applies, find dz/dt when t = 1.?

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