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A hollow sphere of radius r is being filled with water. The surface area of a hemisphere is 3pi*r^2. Question: When the water is at height r, and filling at a rate of 4cm^3s^-1, what is dS/dT?

By the chain rule ds/dt = ds/dr * dr/dv * dv/t. At a height of r, the water fills a hemisphere. So ds/dr = 6pi r. dr/dv = 1/(dv/dr), so we need to find dv/dr. Students should have the formula for the volume ...
HB
Answered by Henry B. • Maths tutor
6187 Views

Solve the equation 5^(2x) - 12(5^x) + 35 = 0

The first step to solving this is equation is to notice that the equation is of a similar to the form of a quadratic equation: ay^2 + by + c = 0 where a, b and c are constants. Next we introduce a new variab...
JG
Answered by Jacob G. • Maths tutor
10947 Views

Differentiate with respect to x: y = ln(x^2+4*x+2).

Let u = x 2 +4x+2 so y = ln(u). Then dy/du = 1/u and du/dx = 2x+4. Using the chain rule we have: dy/dx = (dy/du)*(du/dx) = (1/u)*(2x+4) = (2x+4)/(x 2 +4x+2).
OL
Answered by Okim L. • Maths tutor
5408 Views

Prove that sec^2(θ) + cosec^2(θ) = sec^2(θ) * cosec^2(θ)

These problems can be tricky as they use unfamiliar trigonometric functions such as secant and cosecant. It is much easier to approach these problems by replacing these trigonometric functions with more fami...
HR
Answered by Hugh R. • Maths tutor
12776 Views

find dy/dx at t, where t=2, x=t^3+t and y=t^2+1

We know from simple fraction rules that dy/dx=(dy/dt)/(dx/dt). dy/dt=2t, dx/dt=3t^2+1. Therefore, dy/dx=2x2/12+1=4/13
NO
Answered by Niamh O. • Maths tutor
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