# Tutor profile: Oli H.

## Questions

### Subject: Applied Mathematics

A piece of art is on a fixed horizontal platform. It is ovular in shape, whereby its cross-section may be modelled by the equation x^2 + 2xy + 2y^2 = 10, where x and y are measured in metres. The x and y axes are horizontal and vertical respectively. There is no measuring equipment available and the door to the room in question is 5m but is a long walk. Will the piece of art fit? Tip: The piece of art cannot be assumed to have its base at a height of 0m.

Hint: You need to use every piece of information given. 'Ovular' indicates there are two stationary points. A maximum (at the top) and a minimum (at the bottom). Remember at stationary points the gradient is 0. Therefore we need to find the y value (height value) at each of these points to find the distance between, and hence the height of the piece of art. To find the gradient, we differentiate the equation given. Giving us: 2x + 2y + 2x(dy/dx) + 4y(dy/dx) = 0 At stationary points the gradient terms (dy/dx) become 0. Giving us, 2x + 2y = 0 Dividing by 2 and rearranging we get: x = -y Now we can represent x terms using y terms. Now refer back to the original equation (the height equation) and replace the x terms with y terms. (-y)^2 + 2(-y)y + 2y^2 = 10 With an equation of only y terms, we can solve for the values of y. Simplifying gives: y^2 -2y^2 + 2y^2 = 10 y^2 = 10 Resolving y gives: y = +-(10)^0.5 Remember in calculus (integrating or differentiating) we are dealing with DISPLACEMENTS and not distances. Therefore these values can be positive or negative. Displacements are similar to distances but account for direction. Therefore height = 10^0.5 -- 10^0.5 = 10^0.5 + 10^0.5 = 6.32m Hence the piece of art will not fit through the door as it is taller than 5m.

### Subject: Trigonometry

A lantern is 50m above the sea. The lantern is seen by the Captain as he looks at an angle 30 degrees (assume from the horizontal). 10 seconds later the Captain must adjust his head to 65 degrees in order to directly gaze at the lantern. What speed is the boat travelling?

First thing to do is to draw the problem. (I am not able to insert a diagram). Draw a right angled triangle, height 50m with two angles. The closest angle to the right angle will be 60 degrees, the angle further away will be 30 degrees. This is because as the horizontal increases (the Captain moves further from the balloon in the horizontal plane), the angle from the horizontal will decrease. The triangle you have drawn can be split into two. Ensure a right angle remains within both of these triangles. We will use basic trigonometry rules to resolve this problem. Remembering that tan(angle) = opposite / adjacent Gives us: tan(60) = 50 / x *for the first triangle tan(30) = 50 / y *for the second triangle Where x & y are horizontal distances for each triangles. By multiplying by x or y, and dividing by tan(angle) this gives x = 289 & y = 1374 *Displayed as integers to make clearer, however in practice always display as exact values as this will remove chance of rounding errors. By calculating the difference in horizontal distances, we can calculate the distance the boat has travelled in the time the lantern moved. *Remembering this number is always positive. distance travelled = 1374 - 289 = 1085 speed = distance / time We are given time in the question. speed = 1085/10 = 108.5m/s This would definitely be a speed boat!

### Subject: Calculus

A curve has equation y = x^3 - 48x The point A on the curve has x coordinate -4 The point B on the curve has x coordinate -4+h , where h = height Is A a stationary point?

A stationary point indicates that the gradient at that coordinate is 0. Hence we need to find the gradient of the line, set this to 0, and then work out x and y. Find dy/dx *the equation for the gradient of the curve y. y = x^3 - 48x *Differentiation involves two steps which must be performed in the following order. If they are not performed in this order you WILL get a different result. 1 - Bring the power down to the front 2 - Reduce the power by 1 Hence, dy/dx = 3x^2 - 48 We are told in the question that x = -4 at this point on the curve. Therefore you need to substitute 4 into the equation. Gives: 3(4)^2 - 48 = 0 Always use a calculator and never do maths in your head, even if you are confident you will get it right. The total number of left and right brackets must be equal. Gradient = 0 at this point verifies that A is a stationary point.

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