A point moves along the graph y= x^(5/2) so that its x coordinate increases at the constant rate of 2units/second. Find the rate at which its y coordinate is increasing as it passes the point (4 ,32)?

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Answer 1 · 2015-08-26T14:12:57

$\frac{d y}{d t} = 40$ $dy/dt = 40$ ( $y$ $y$ -units)/second

If $y = x^{\frac{5}{2}}$ $y= x^(5/2)$

and $\frac{d x}{d t} = 2$ $dx/dt = 2$ ( $x$ $x$ -units)/second.

find $\frac{d y}{d t}$ $dy/dt$ when $x = 4$ $x=4$ (and $y = 32$ $y=32$ ).

Differentiate $y = x^{\frac{5}{2}}$ $y= x^(5/2)$ with respect to $t$ $t$ (use implicit differentiation):

$\frac{d y}{d t} = \frac{5}{2} x^{\frac{3}{2}} \frac{d x}{d t}$ $dy/dt= 5/2x^(3/2) dx/dt$

Substitue the known value for $x$ $x$ and $\frac{d x}{d t}$ $dx/dt$ and solve for $\frac{d y}{d t}$ $dy/dt$ :

$\frac{d y}{d t} = \frac{5}{2} x^{\frac{3}{2}} \frac{d x}{d t}$ $dy/dt= 5/2x^(3/2) dx/dt$

$= \frac{5}{2} {(4)}^{\frac{3}{2}} (2)$ $= 5/2(4)^(3/2) (2)$

$= 40$ $= 40$