An industrial tank is used to store liquid chemicals. The tank is cylindrical with a total height of 210 centimeters. The liquid is drained through an outlet valve located 10 centimeters above the base of the tank. At time ttt hours after the valve is opened, the depth of liquid, hhh centimeters, decreases at a rate which is proportional to h−10h - 10h−10.
Initially, the tank is completely full, and the depth of the liquid is decreasing at a rate of 4 centimeters per hour.
Show that
dhdt=−0.02(h−10) \frac{dh}{dt} = -0.02(h - 10) dtdh=−0.02(h−10)Solve the differential equation
dhdt=−0.02(h−10) \frac{dh}{dt} = -0.02(h - 10) dtdh=−0.02(h−10)to find an expression for hhh in terms of ttt.
Find the time taken for the depth of the liquid to reach 110 centimeters. Give your answer to the nearest hour.
425 exam-style questions on OCR A Level Maths 1.7 Differentiation, covering 1.7.1 Derivative as gradient of the tangent, 1.7.2 Gradient of the tangent at a point, 1.7.3 Sketching the gradient function, 1.7.4 Second derivatives, 1.7.5 Second derivative as rate of change of gradient, 1.7.6 Convex, concave and points of inflection (A-level only), 1.7.7 Differentiation from first principles for powers of x, 1.7.8 Differentiation from first principles for sin x and cos x (A-level only), 1.7.9 Differentiating x^n, 1.7.10 Differentiating e^(kx) and a^(kx) (A-level only), 1.7.11 Differentiating trigonometric functions (A-level only), 1.7.12 Derivative of ln x (A-level only), 1.7.13 Tangents and normals, 1.7.14 Stationary points, 1.7.15 Increasing and decreasing functions, 1.7.16 Points of inflection (A-level only), 1.7.17 Product and quotient rules (A-level only), 1.7.18 Chain rule (A-level only), 1.7.19 Parametric and implicit differentiation (A-level only), 1.7.20 Constructing differential equations (A-level only), and 1.7 Differentiation. Each one has a worked solution and a mark scheme showing where the marks go.