The height of a cylinder is increasing at a constant rate of 10 meters per minute, and the volume is increasing at a rate of 1135 cubic meters per minute. At the instant when the height of the cylinder is 9 meters and the volume is 354 cubic meters, what is the rate of change of the radius? The volume of a cylinder can be found with the equation V=πr2h. Round answer to three decimal places

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pattererX · Accepted Answer

Given: height of a cylinder is increasing at a constant rate of 10m/min. i.e dhdt (h: height of cylinder) Volume is increasing at a rate of 1135 cubic meters/min. dVdt=1135 m3/min At the instant V=354m3 h=9 m Relation between volume (V), radius (r) and height (h) V=πr2h Differentiate above with respect ot time [dVdt=ddt(πr2h)] (Take constant utside and apply product rule) [⇒dVdt=πddt(r2h) ⇒dVdt=πbigg[r2dhdt+hd(r2)dtbigg] apply chain rule (above) [⇒dVdt=πbigg[r2dhdt+hd(r2)dt×drdhbigg] ⇒dVdt=πbigg[r2dhdt+h2rdrdtbigg]bigg(ddx(xn)=nxn−1bigg) ⇒dVdt=πr2dhdt+2πrhdrdt] Given: [dhdt=10 mmin dVdt=1135 m3min At the instant: V=354 i.e [πr2h=354 m3 h=9m

The height of a cylinder is increasing at a constant rate of 10 meters

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