Eric Nicholas K's answer to Georgia Jansen's Secondary 4 A Maths Singapore question.
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Part (i) requires similar triangles.
Part (ii) requires you to find dh/dt. We are given dV/dt, so we need to find dV/dh, which comes from the first part equation V = 3h2.
However, to find the numerical value of dV/dh after differentiating V, we need a specific value of h.
We have to obtain this by using the fact that dV/dt is given and the water is running for a fixed time to evaluate V and ultimately obtain a value for h.
From there, we use the chain rule equation dV/dt = dV/dh dh/dt to solve for dh/dt.
Part (ii) requires you to find dh/dt. We are given dV/dt, so we need to find dV/dh, which comes from the first part equation V = 3h2.
However, to find the numerical value of dV/dh after differentiating V, we need a specific value of h.
We have to obtain this by using the fact that dV/dt is given and the water is running for a fixed time to evaluate V and ultimately obtain a value for h.
From there, we use the chain rule equation dV/dt = dV/dh dh/dt to solve for dh/dt.
Date Posted:
5 years ago
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